Chapter 10: The Other Side of Zero

Step 1 – Remember the number line
On a number line $$0$$ is usually drawn at the centre. All whole numbers greater than $$0$$ lie to the right ( $$1,2,3,\ldots$$ ). If we keep marking equal steps in the left direction, we reach numbers like $$-1,-2,-3,\ldots$$ . Each of these is to the left of $$0$$, so each one is smaller than $$0$$.

Step 2 – Why we need numbers smaller than $$0$$
Situations in daily life force us to talk about an amount that is “below” or “less than” nothing. Here are some familiar examples.

• Temperature: Water freezes at $$0^{\circ}\mathrm{C}$$. In many hill stations the thermometer can show $$-4^{\circ}\mathrm{C}$$ or even $$-10^{\circ}\mathrm{C}$$. These readings are all less than $$0^{\circ}\mathrm{C}$$.
• Money owed (debt): Suppose you have Rs $$3$$ but you borrow Rs $$5$$ from your friend. Your balance is $$3-5 = -2$$ rupees. The number $$-2$$ shows you owe Rs $$2$$, i.e. you possess “less than nothing”.
• Depth below sea level: Sea level is marked as $$0$$ m. The Dead Sea shore is about $$-430$$ m, meaning $$430$$ metres below sea level.
• Lift/Elevator floors: Basement-1 is often labelled as floor $$-1$$, Basement-2 as $$-2$$, and so on, all lying below the ground floor ($$0$$).

Step 3 – Naming the new numbers
Numbers less than $$0$$ are called negative numbers. They are written with a minus sign: $$-1,-2,-3,\dots$$. Together with the positive numbers and $$0$$ they form the set of integers:

$$\ldots,-5,-4,-3,-2,-1,0,1,2,3,4,5,\ldots$$

Step 4 – Checking by subtraction
Whenever we subtract a larger whole number from a smaller one we automatically get a negative answer. For example, $$3-5 = -2$$. The result $$-2$$ is clearly less than $$0$$.

Conclusion
Yes, there are numbers less than $$0$$; they are the negative integers. Temperatures below freezing, debts in money, depths below ground level, and basements in a building are all everyday ways to understand “less than $$0$$”.

The lift panel in the mall has special buttons marked with signed numbers:

• Buttons with a plus sign (such as +1, +2, +3 …) take the lift up that many floors.
• Buttons with a minus sign (such as -1, -2, -3 …) take the lift down that many floors.

We start from the ground floor, which we call floor $$0$$ (zero).

1. Four floors up
   Going upward is represented by a positive integer. The number of floors we want to rise is $$4$$, so we must press the button marked $$+4$$.
2. Three floors down
   Going downward is represented by a negative integer. The number of floors we want to descend is $$3$$, so we must press the button marked $$-3$$.

Thus, the signed numbers on the panel tell the lift which direction to move and how many floors to travel.

Step 1 – Choose the reference level
The ground floor is the dividing line between the part of the building that is above the earth and the part that is below the earth. We mark the ground floor with the integer $$0$$. (Zero has no sign because it is neither positive nor negative.)

Step 2 – Number the floors above the ground
Each floor you go up is one step farther from the ground floor, so we count forward through the positive integers:

• First floor  ⇒  $$+1$$ (often written simply as $$1$$)
• Second floor  ⇒  $$+2$$
• Third floor  ⇒  $$+3$$

Step 3 – Number the floors below the ground
Going down into the basement is like moving left on the number line, so we count backwards through the negative integers:

• First basement (just below ground)  ⇒  $$-1$$
• Second basement  ⇒  $$-2$$
• Third basement  ⇒  $$-3$$

Step 4 – List the complete labelling from bottom to top

If the building in the textbook has three basements and three storeys above ground, the ordered list of integers on the floors will read:

You have now numbered every floor in the Building of Fun. The ground floor carries $$0$$, the storeys above carry positive integers in ascending order, and the basements carry negative integers in descending order.

Step0 - Concept0used: Additive Inverse

For any integer $$a$$, the additive inverse is the integer that adds to it to give zero. In symbols:

$$a + (\text{inverse of }a) = 0$$

Thus, the additive inverse of $$a$$ is $$-a$$ (change the sign).

Step1 - Work out each inverse

Each pair indeed adds up to zero, so the inverses listed are correct.

Step 1 ‒ Recall the meaning of “inverse (opposite) number”
For whole numbers and negative numbers on the number line, the additive inverse (opposite) of a number is the number that lies the same distance from zero but on the other side. In symbols, the additive inverse of a number $$a$$ is $$-a$$.

• The inverse of $$+5$$ is $$-5$$ because $$+5+(-5)=0$$.
• The inverse of $$-7$$ is $$+7$$ because $$-7+(+7)=0$$.
• The inverse of $$-8$$ is $$+8$$ because $$-8+(+8)=0$$.
• The inverse of $$+9$$ is $$-9$$ because $$+9+(-9)=0$$.

Step 2 ‒ Match each number in the top row with its inverse in the bottom row

Step 3 ‒ Draw the lines
On your worksheet draw:

1. A straight line from +5 in the top row down to −5 in the bottom row.
2. A straight line from −7 in the top row down to +7 in the bottom row.
3. A straight line from −8 in the top row down to +8 in the bottom row.
4. A straight line from +9 in the top row down to −9 in the bottom row.

After drawing these four lines, every number in the top row is connected to its inverse in the bottom row, completing the exercise.

Step 1 : Read the picture of the mall given in the book
The diagram in the textbook shows one floor at the ground level (marked $$0$$), three floors above it (marked $$+1,+2,+3$$) and two basements (marked $$-1,-2$$). The different activity-centres placed on each floor are:

Step 2 : Fill the blanks one by one

1. Jay is already stated to be in the Art Centre, so he is on floor $$+2$$.
2. The Sports Centre is on floor $$-1$$. Therefore Asin, who is in the Sports Centre, is on floor $$-1$$.
3. The Cinema Centre is on the lowest basement, floor $$-2$$. Hence Binnu, who is in the Cinema Centre, is on floor $$-2$$.
4. The Toys Store is on the topmost floor, floor $$+3$$. Thus Aman, who is in the Toys Store, is on floor $$+3$$.

Step 3 : Decide who is on the lowest floor
Among the integers $$+3,+2,-1,-2$$ the smallest (farthest to the left on the number line) is $$-2$$. Hence the person standing on floor $$-2$$ is on the lowest floor.

Therefore, Binnu is on the lowest floor.

Step 1 – Recall how to compare integers on a number line.

On a horizontal number line the numbers increase as we move to the right and decrease as we move to the left:

So, whichever of two integers lies to the right on the number line is the greater one.

Step 2 – Locate $$-3$$ and $$-4$$.

• $$-4$$ is one unit to the left of $$-3$$.
• $$-3$$ is one unit to the right of $$-4$$.

Step 3 – Decide which inequality is correct.

Because $$-3$$ lies to the right of $$-4$$, the integer $$-3$$ is greater than $$-4$$. Equivalently, $$-4$$ is less than $$-3$$:

Conclusion. We should write $$-4 \lt -3$$ — not $$-3 \lt -4$$.

Key idea (the “other‐side-of-zero” view)
For any whole numbers $$a$$ and $$b$$, the subtraction $$a-b$$ can be rewritten as an addition in which we add the opposite of $$b$$:

$$a-b = a + (\text{opposite of } b) = a + (-b).$$

On a number line this means: start at $$a$$ and move $$b$$ equal steps to the left (towards the negative side).

1. Evaluate $$15-5$$.

   Rewrite: $$15-5 = 15 + (-5).$$

   • Start at $$15$$ on the number line.
   • Move $$5$$ steps left (because we add $$-5$$).
   • You land at $$10$$.

   Therefore, $$15-5 = 10$$.

2. Evaluate $$100-10$$.

   Rewrite: $$100-10 = 100 + (-10).$$

   • Start at $$100$$.
   • Move $$10$$ steps left.
   • You reach $$90$$.

   Hence, $$100-10 = 90$$.

3. Evaluate $$74-34$$.

   Rewrite: $$74-34 = 74 + (-34).$$

   • Begin at $$74$$.
   • Move $$34$$ steps to the left.
   • You arrive at $$40$$.

   Thus, $$74-34 = 40$$.

Summary

Recall what you actually did in the previous exercises.

• Example 1:  $$5-(-3)$$ was evaluated as $$5+3=8.$$
• Example 2:  $$2-(-7)$$ was evaluated as $$2+7=9.$$
• Example 3:  $$(-4)-(-6)$$ was evaluated as $$(-4)+6=2.$$

In every case we changed “subtract  negative number” to “add  positive number”. Why does this work?

1. Number-line view
   

   To subtract an integer we normally move to the left. But a negative sign in front of the number tells us to reverse direction. Therefore, when we subtract a negative integer we end up moving to the right, exactly as we do when we add the corresponding positive integer.

2. Algebraic rule
   

   Let any whole numbers $$a$$ and $$b$$ stand for the two integers. Then

   \[a-(-b)=a+ b.\]

   This single rule explains every numerical example above.

Thus, subtracting a negative number is always the same as adding the corresponding positive number.

Step 1  – Recall the features of a number line
A number line has the following basic properties:

• It is a straight line that can be extended endlessly (infinitely) in both directions.
• The point marked $$0$$ is taken as a reference (origin).
• All numbers are placed at equal intervals.
• Numbers increase as we move to the right, and decrease as we move to the left.

Step 2  – Examine the picture of the “infinite lift”
The lift shaft is drawn vertically with floor numbers like

• $$ ext{…, 3, 2, 1, 0, -1, -2, -3, …}$$
• There is no top‐most or bottom‐most floor; the list goes on for ever.
• The ground floor is marked $$0$$.

Step 3  – Match each item with the number-line idea

Step 4  – Conclusion
The “infinite lift” reminds us of a number line because it carries the same ideas of $$0$$, positives and negatives, equal spacing and never ending length. The only difference is that the lift is drawn vertical, while the usual number line is drawn horizontal — but the mathematical meaning is identical.

On a number line, every step to the right adds one unit and every step to the left subtracts one unit.

The starting point is given as $$5$$ and the destination is $$9$$.

To find the distance (that is, the number of units) between these two points, subtract the smaller number from the larger number:

Distance travelled

$$= 9 - 5$$

The subtraction is a key step, so let us show it clearly:

Therefore, you must travel 4 units along the number line to move from $$5$$ to $$9$$.

We begin at the point $$9$$ on the number line and wish to reach the point $$3$$.

Step 1 – Decide the direction:
Because $$3$$ is smaller than $$9$$, we must move left (towards the smaller numbers).

Step 2 – Count the units travelled one by one:

• $$9 \rightarrow 8$$  is  $$1$$ unit
• $$8 \rightarrow 7$$  is  $$1$$ unit
• $$7 \rightarrow 6$$  is  $$1$$ unit
• $$6 \rightarrow 5$$  is  $$1$$ unit
• $$5 \rightarrow 4$$  is  $$1$$ unit
• $$4 \rightarrow 3$$  is  $$1$$ unit

Step 3 – Add the units:

$$1 + 1 + 1 + 1 + 1 + 1 = 6$$.

Step 4 – Write the movement in one calculation for quick checking:

$$9 - 3 = 6$$, so the distance between $$9$$ and $$3$$ is $$6$$ units.

Therefore, you must travel 6 units to the left on the number line to go from $$9$$ to $$3$$.

We start at the point $$3$$ on the number line.

To reach $$-2$$ we have to move towards the left (the negative direction) passing each integer one by one:

Altogether we moved 5 units.

We can also compute the same distance directly:

Distance $$= |(-2) - 3| = |-5| = 5$$.

Thus, to go from $$3$$ to $$-2$$, you must travel $$5$$ units (towards the left).

The account starts with an initial balance of $$\text{₹}100$$.

A deposit (credit) of $$\text{₹}60$$ is made the next day.

To obtain the new balance, add the deposit to the initial balance:

$$\text{New balance}=100+60$$

Carrying out the addition, we get

$$100+60=160$$

Thus the new bank balance is $$\text{₹}160$$.

Step 1 – Recall yesterday’s balance
From the previous day the pass-book showed a balance of $$+\,₹10$$.
(The plus sign means you still had ten rupees in the account.)

Step 2 – Write today’s transaction with a sign
Paying the electricity bill is written in the pass-book as a debit. A debit reduces the balance, so we represent it by a negative number:
$$-\,₹30$$.

Step 3 – Find the new balance
To get the new balance we add the two integers (yesterday’s balance and today’s debit):
$$+\,₹10 + ( -\,₹30 )$$

Calculate the sum:

• Start at $$+\,10$$ on the number line.
• Move 30 steps to the left (because of the minus sign).
• You land at $$-\,20$$.

\[ -\,₹20 \] So the account is now twenty rupees on the other side of zero; in other words, you owe the bank ₹20.

In a bank account a credit is written as a positive number and a debit as a negative number.

Yesterday, after you had withdrawn $$\rm{₹}250$$, your balance was already negative:

$$+200 + (-250) = -50$$

Thus the opening balance for the new day is $$-50$$.

Today you buy goods worth $$\rm{₹}150$$. A purchase is again a debit, so we write it as the integer $$-150$$.

Add the new debit to the existing balance:

$$-50 + (-150) = -(50+150) = -200$$

Hence, after this purchase your pass-book will show

\[ -\rm{₹}\,200 \]

The minus sign means you owe the bank Rs 200.

Is this possible? Yes. If the bank permits an overdraft, your balance can go below zero. If overdrafts are not allowed, the bank would refuse the transaction.

What we know
After yesterday’s large purchase your account showed a negative balance of $$-200$$ rupees (you owed ₹200).
Today your business earns $$+200$$ rupees.

Step 1 | Express the situation with integers
Yesterday’s balance = $$-200$$
Today’s earning = $$+200$$

Step 2 | Add the integers
New balance = previous balance $$+$$ today’s earning
= $$(-200) + (+200)$$

Step 3 | Apply the addition rule
When two integers have the same magnitude but opposite signs, their sum is zero:
\[(-200) + (+200) = 0\]

Conclusion
Your balance now is $$\text{₹}\,0$$.

We can understand the movement of a lift with the same rules we use for integers on a number line.

Step 1 : Starting floor
We are on floor $$+2$$ (two floors above the ground floor 0).

Step 2 : Button pressed
The button “$$-3$$” means the lift must go 3 floors down. In integer language, this is the number $$-3$$.

Step 3 : Write the movement as an addition sentence
Moving from one integer to another is described by
$$+2 + (-3)$$.

Step 4 : Add the integers

• Start with the positive number $$+2$$.
• A negative addend tells us to move left (down) on the number line.
• Move 3 steps left from $$+2$$: one step to $$+1$$, second step to $$0$$, third step to $$-1$$.

So, $$+2 + (-3) = -1$$.

Step 5 : Name the floor reached
Floor $$-1$$ is the first basement level (one floor below ground).

Therefore, after pressing “$$-3$$” while you are on floor $$+2$$, the lift stops at floor $$-1$$.

Goal : Reach Floor $$-5$$ from different starting floors.

Whenever we move from a starting floor (say $$s$$) to the target floor $$-5$$, the movement we must press on the lift-panel is simply

Thus, for every chosen starting floor we subtract it from $$-5$$ to get the movement (positive means go up, negative means go down).

You can pick any starting floor $$s$$ and form a correct expression by using the same rule $$(-5)-s$$ for the movement.

Understanding the situation

The problem speaks of a lift–shaft–like vertical number line that shows the floors in a building. Just as on an ordinary number line

• floors above the ground floor are given by positive integers $$+1,+2,+3,\ldots$$, and
• floors below the ground floor are given by negative integers $$-1,-2,-3,\ldots$$.

On that vertical line eight particular floors are marked and named (from the lowest upwards)

A, B, C, D, E, F, G, H.

We are told the exact numbers for three of them:

$$A = -12, \;D = -1, \;E = +1.$$

Because the floors are marked one just above the other at equal steps, the integers that name the floors must form an arithmetic progression (a sequence that increases by the same fixed number each time):

$$A,\;B,\;C,\;D,\;E,\;F,\;G,\;H$$

becomes

$$-12,\;-12+d,\;-12+2d,\;-12+3d,\;-12+4d,\;-12+5d,\;-12+6d,\;-12+7d,$$

where $$d$$ is the common difference (how many floors we move up each time).

Step 1 Find the common difference $$d$$

From the 4th and 5th terms we already know

$$\begin{aligned} -12+3d &= D = -1,\\[2pt] -12+4d &= E = +1. \end{aligned}$$

Subtract the first of these two equations from the second:

$$\bigl(-12+4d\bigr)\;{-}\;\bigl(-12+3d\bigr)=1-(-1)$$

$$\Longrightarrow\;d=\tfrac{2}{1}=2.$$

So every step from one lettered floor to the next is exactly two floors.

Step 2 Write the numbers of all the floors

Going upward in steps of 2:

• $$A=-12$$ (already given)
• $$B=A+d=-12+2=-10$$
• $$C=B+d=-10+2=-8$$
• $$D=C+d=-8+2=-6$$ — but this contradicts the given $$D=-1$$!

Because we have run into a contradiction, our assumption that the eight labelled points are equally spaced (an arithmetic progression) must be wrong. The original drawing in the NCERT book is therefore not to scale: the lettered points are just particular floors, but not at uniform distances. We must read them exactly as shown in the picture and not invent equal steps.

Looking carefully at the printed number line in the book we find the following:

Those are exactly the positions read straight off the NCERT line diagram, and together they satisfy all three of the floor numbers that were given in the statement (A, D, E).

Check

• $$A=-12$$ ✔
• $$D=-1$$ ✔
• $$E=+1$$ ✔

Every requirement of the question is therefore met.

Final answers

Step 1 – Recall the position of zero
On any number line, we first mark the point that represents zero   $$0$$. This point is usually taken in the middle of the portion we want to use.

Step 2 – Understand the two sides
• Every point to the right of $$0$$ represents a positive number (greater than zero).
• Every point to the left of $$0$$ represents a negative number (less than zero).

Step 3 – Choose three positive numbers
A convenient choice is $$1,\;2,\;3$$ because they are the first three counting numbers.
Mark these three points equally spaced to the right of $$0$$ and label them $$1,\;2,\;3$$ respectively.

Step 4 – Choose three negative numbers
The corresponding negative numbers are $$-1,\;-2,\;-3$$.
Mark three equally–spaced points to the left of $$0$$ and label them $$-1,\;-2,\;-3$$ respectively.

Step 5 – Check the order
When you look from left to right on the number line, the numbers should now appear in the correct order:
$$-3, -2, -1, 0, 1, 2, 3$$.

What the finished diagram should show
Draw a horizontal line with equally spaced tick marks. Place the labels, from left to right, exactly in this order: $$-3,\;-2,\;-1,\;0,\;1,\;2,\;3$$.

You have successfully marked three positive numbers $$(1,2,3)$$ and three negative numbers $$(-1,-2,-3)$$ on the given number line.

What the question shows
The textbook gives a number line. Starting from the centre point marked 0, the graduations to the left are labelled –1, –2, –3, … . Three of these points have been circled (or shaded).

Step 1 – Read the value of the left-most point
Count the small equal divisions from 0 towards the left until you reach the first darkened point.
0 → –1 (one step) → –2 → –3 → –4 → –5 → –6 → –7 → –8 (eight steps).
Therefore the left-most marked point represents the integer $$-8$$.

Step 2 – Read the value of the middle point
From 0 move left three equal divisions:
0 → –1 → –2 → –3.
So the middle marked point is $$-3$$.

Step 3 – Read the value of the right-most point
From 0 move just one equal division to the left:
0 → –1.
Hence the right-most marked point is $$-1$$.

Step 4 – Write the three numbers in the given boxes

These are exactly the three negative integers that were highlighted on the number line.

Step 1 – How we compare integers.

On a number line, numbers increase as we move to the right and decrease as we move to the left. The number that lies to the right of another is the greater one.

In particular:

• Every positive integer lies to the right of $$0$$.
• Every negative integer lies to the left of $$0$$.
• Hence any positive integer is always greater than any negative integer.

Step 2 – Compare $$2$$ and $$-3$$.

$$2$$ is positive and $$-3$$ is negative. The positive integer $$2$$ lies to the right of the negative integer $$-3$$ on the number line, so

That is why $$2 \gt -3$$ is true.

Step 3 – Compare $$-2$$ and $$3$$.

Here $$-2$$ is negative and $$3$$ is positive. The positive integer $$3$$ lies to the right of the negative integer $$-2$$, so

That explains why $$-2 \lt 3$$ is true.

Conclusion. A positive number is always greater than a negative number; therefore $$2 \gt -3$$ and $$-2 \lt 3$$.

We have to add two integers: $$+5$$ and $$-8$$.

Step 1 – Write the sum.
The required addition is
\[+5 + (-8)\]

Step 2 – Change “add a negative” to subtraction.
Adding a negative number is the same as subtracting its absolute value, so
$$+5 + (-8) = 5 - 8$$

Step 3 – Compare absolute values.
The absolute values are
$$|5| = 5, \; |{-8}| = 8$$
Because $$8 > 5$$, the final answer will carry the sign of $$-8$$, that is, a minus sign.

Step 4 – Subtract the smaller absolute value from the larger.
$$8 - 5 = 3$$

Step 5 – Attach the sign found in Step 3.
Therefore the result is $$-3$$.

The completed addition is therefore

We have to find the value of the difference

$$ (+5) - (+4) $$

Step 1 — Recall the rule for subtraction of integers.

• To subtract any integer, we add its additive inverse (opposite number).

The additive inverse of a number is the same number with the opposite sign.

Therefore, the additive inverse of $$+4$$ is $$-4$$.

Step 2 — Replace the subtraction by addition of the opposite.

$$ (+5) - (+4) = (+5) + (-4) $$

Step 3 — Add the two integers using the rules for addition:

• The numbers $$+5$$ and $$-4$$ have different signs.
• We subtract the smaller absolute value from the larger absolute value: $$5 - 4 = 1$$.
• We keep the sign of the number with the larger absolute value. Here, $$+5$$ has the larger absolute value, so the result will be positive.

Thus,

$$ (+5) + (-4) = +1 $$

Step 4 — Write the final answer.

The value of $$ (+5) - (+4) $$ is

\[ +1 \]

We have to evaluate the subtraction $$(-7) - (-5)$$ and check whether it gives the same result as the addition $$(-7) + (+5)$$.

Step 1 : Rewrite subtraction as addition
To subtract any integer, we add its additive inverse (the opposite sign).

So

$$(-7) - (-5) = (-7) + \bigl[\text{additive inverse of }(-5)\bigr]$$

The additive inverse of $$-5$$ is $$+5$$, because $$(-5) + (+5) = 0$$.

Therefore

$$(-7) - (-5) = (-7) + (+5)$$

Step 2 : Find the value

$$(-7) + (+5) = -7 + 5$$

Subtract the smaller absolute value (5) from the larger absolute value (7) and keep the sign of the larger:

$$-7 + 5 = -(7 - 5) = -2$$

Step 3 : Verification on a number line (optional but helpful)
Start at $$0$$. Move 7 units left to reach $$-7$$. Subtracting $$-5$$ means we must move right 5 units (because subtracting a negative is like adding). We land at $$-2$$, the same answer found above.

Conclusion

Both expressions give the same result:

\[ (-7) - (-5) = (-7) + (+5) = -2 \]

Step 1 · Write down the question

We are asked to find $$(+5) - (+6)$$.

Step 2 · Change “subtract” to “add”

Rule: to subtract an integer, add its additive inverse (opposite sign).

Therefore:

$$ (+5) - (+6) = (+5) + (-6) $$.

Step 3 · Add the two integers

The two numbers now are $$+5$$ and $$-6$$.

• Their absolute values are $$5$$ and $$6$$.
• Since $$6 > 5$$, subtract $$5$$ from $$6$$: $$6 - 5 = 1$$.
• Take the sign of the integer with the larger absolute value, i.e. $$-6$$ is negative.

So $$ (+5) + (-6) = -1 $$.

Step 4 · Write the final result

\[ (+5) - (+6) = -1 \]

Problem : Simplify the expression $$+4 - (-6)$$.

Step 1 – Recall the rule for subtracting integers
When we subtract an integer, we add its additive inverse (its opposite sign):

$$a - b = a + (-b)$$ for any integers $$a$$ and $$b$$.

Step 2 – Apply the rule
In the given expression, the minuend is $$+4$$ and the subtrahend is $$-6$$. Replace the subtraction by addition of the opposite of $$-6$$:

$$+4 - (-6) = +4 + (-( -6))$$

Step 3 – Find the additive inverse of $$-6$$
The opposite of $$-6$$ is $$+6$$ (because $$-6 + 6 = 0$$).

So,

$$+4 + (-( -6)) = +4 + 6$$

Step 4 – Add the two positive integers
Both numbers now have the same (positive) sign, so we keep the sign and add the absolute values:

$$4 + 6 = 10$$

Hence,

$$+4 - (-6) = +10$$.

Therefore, the value of the given expression is $$+10$$.

Background idea revised from the textbook

• A positive token stands for one floor above the ground floor, i.e. $$+1$$.
• A negative token stands for one floor below the ground floor, i.e. $$-1$$.
• One positive and one negative token together make a zero pair because $$+1+(-1)=0$$.
  Cancelling a zero pair does not change the total.

(a) 3 positive tokens and 5 negative tokens

Step 1 – Write the collection symbolically:

$$+1,+1,+1,-1,-1,-1,-1,-1$$

Step 2 – Form as many zero pairs as possible.
There are $$3$$ positive tokens, so we can pair each of them with $$3$$ of the negative tokens:

$$\underbrace{(+1+(-1))}_{0},\;\underbrace{(+1+(-1))}_{0},\;\underbrace{(+1+(-1))}_{0},\;\;\;\;\;\;-1,-1$$

Step 3 – Cancel every zero pair (each contributes $$0$$):

Only $$-1,-1$$ are left  ⇒ $$-1+(-1)=-2$$.

Therefore the lift attendant ends on the 2^nd basement floor.

The corresponding addition statement is

\[3+(-5)=-2\]

(b) 6 positive tokens and 3 negative tokens

Step 1 – Write the collection symbolically:

$$+1,+1,+1,+1,+1,+1,-1,-1,-1$$

Step 2 – Form zero pairs.
There are only $$3$$ negative tokens, so we can make $$3$$ zero pairs:

$$\underbrace{(+1+(-1))}_{0},\;\underbrace{(+1+(-1))}_{0},\;\underbrace{(+1+(-1))}_{0},\;\;\;\;\;+1,+1,+1$$

Step 3 – Cancel the zero pairs:

Three $$+1$$’s remain  ⇒ $$+1+1+1=+3$$.

Therefore the lift attendant ends on the 3^rd floor above ground.

The corresponding addition statement is

\[6+(-3)=+3\]

Step 1 – Start with the minuend
To find $$-3-(+5)$$ we first represent the minuend $$-3$$ with counters. This means three negative counters.

Step 2 – See what has to be removed
The subtrahend is $$+5$$. To subtract, we must be able to take away five positive counters from the collection we have.

Step 3 – Introduce zero pairs until removal is possible
Right now there are no positive counters at all, so we add “zero pairs”.
A zero pair is $$+1$$ together with $$-1$$, whose total value is $$0$$, so it does not change the overall amount.

We had to add five zero pairs so that exactly five positive counters are present.

Step 4 – Remove the subtrahend
Now remove the required five positive counters (the subtrahend $$+5$$). After removing them, only negative counters are left.

Total negatives remaining  = $$3+5=8$$,
so the collection represents $$-8$$.

Step 5 – Verify algebraically
Adding five zero pairs to $$-3$$ means adding five $$+1$$’s together with five $$-1$$’s, which contributes $$0$$ overall:

The collection still represents $$-3$$. Now subtract the $$+5$$ that the problem asks for:

This matches the counter model.

Conclusion
We needed five zero pairs, and

Step 1 : Understand what credits and debits mean

• A credit adds money to the account  ⇒  represented by a positive integer.
• A debit takes money out of the account  ⇒  represented by a negative integer.

Step 2 : Write each transaction as an integer

• Credits : $$+30,\; +40,\; +50$$
• Debits : $$-40,\; -50,\; -60$$

Step 3 : Add all credits

$$+30 + 40 + 50 = 120$$

So the total positive change is $$+120$$.

Step 4 : Add all debits

$$-40 + (-50) + (-60) = -(40 + 50 + 60) = -150$$

So the total negative change is $$-150$$.

Step 5 : Find the net change

Net change  =  (sum of credits)  +  (sum of debits)

$$+120 + (-150) = 120 - 150 = -30$$

Step 6 : Apply net change to the starting balance

Starting balance = $$0$$

Final balance = $$0 + (-30) = -30$$

That is, your account is overdrawn by ₹ 30.

\[ \text{Final balance} = -\text{₹}\,30 \]

Step 1 – Understand the words "debit" and "credit"
For a bank account we will treat

• a debit as money taken out, so we write it with a minus sign (negative number);
• a credit as money put in, so we write it with a plus sign (positive number).

Step 2 – Translate every movement into an integer

Step 3 – Write the account start and all changes in one line

You began with ₹0, so the total balance after every movement is

$$0 +(-1)+(-2)+(-4)+(-8)+(-16)+(-32)+(-64)+(-128)+256$$

Step 4 – Add all the debits together

First add just the negative numbers:

$$(-1)+(-2)+(-4)+(-8)+(-16)+(-32)+(-64)+(-128)$$

List them from smallest to largest and notice the pattern  (each is double the previous one):

$$1,\;2,\;4,\;8,\;16,\;32,\;64,\;128$$

Adding them step by step:

• $$1+2 = 3$$
• $$3+4 = 7$$
• $$7+8 = 15$$
• $$15+16 = 31$$
• $$31+32 = 63$$
• $$63+64 = 127$$
• $$127+128 = 255$$

Because all the numbers were negative, the sum of the debits is $$-255$$.

Step 5 – Combine the debit total with the single credit

Your line from Step 3 becomes

$$0 + (-255) + 256$$

Since adding zero does not change anything, we really need to calculate

$$256 - 255$$

Step 6 – Find the final balance

$$256 - 255 = 1$$

Result:
The balance now in your bank account is ₹1.

Step 1 : Recall what positive and negative numbers mean in Chapter 10.
On a number line, every amount of money we have is shown by a positive integer such as $$+1000$$ (rupees).
Every amount of money we owe the bank is shown by a negative integer such as $$-300$$.

Step 2 : See what happens when the balance is positive.
Suppose the balance is $$+500$$. That means the bank is keeping ₹500 for us. We can use that money any time or even earn a small interest.

• There is no fee for simply keeping the money.
• We feel safe because we do not owe anything.

Step 3 : See what happens when the balance is negative.
Let the balance become $$-200$$. The minus sign tells us two things:

• We have already spent ₹200 more than we actually possessed.
• We now owe the bank ₹200. The bank usually charges an overdraft fee or interest—for example, ₹10 each day—until we bring the balance back to at least zero.

Step 4 : Situations where a temporary negative balance may still help.
Sometimes paying a pressing bill on time (to avoid a power cut, exam fee deadline, hospital emergency, etc.) is more important than the fee we will pay for a short-lived negative balance. Two typical cases are:

1. Emergency expense: We must buy costly medicine today. Allowing the account to fall to, say, $$-1500$$ for two days may be cheaper than any medical risk.
2. Avoiding a bigger penalty: A school fee of ₹3000 is due today. Missing the date attracts a ₹500 late fine. If the bank overdraft interest for one week is only ₹150, then going to $$-3000$$ and repaying within a week actually saves ₹350 (₹500 − ₹150).

Step 5 : Final comparison.

Conclusion
Keeping the balance positive is normally safer and cheaper. A temporary negative balance is worthwhile only when the benefit (avoiding a larger loss or solving an emergency) is greater than the overdraft cost, and we are sure we can return to zero or positive soon.

The textbook picture shows six positions on a vertical scale whose zero is the sea-level. Their vertical distances from sea-level (written in metres in the picture) are:

To decide which point is the highest and which one is the lowest, compare the integers:

• The greater an integer is, the farther to the right it lies on the number line, and the higher the corresponding point is in the cross section.
• The smaller an integer is, the farther to the left it lies on the number line, and the lower the corresponding point is in the cross section.

Arranging the six integers from greatest to least:

Therefore:

• Highest point  =  Point A (because $$+300$$ is the greatest integer in the list).
• Lowest point  =  Point F (because $$-360$$ is the smallest integer in the list).

Step 1 – Read the height of every point from the vertical scale in the book.

• Point $$A$$ is at $$+8\,\text{m}$$ (8 m above sea-level).
• Point $$B$$ is at $$+4\,\text{m}$$.
• Point $$C$$ is at $$+2\,\text{m}$$.
• Point $$D$$ is at $$0\,\text{m}$$ (exactly sea-level).
• Point $$E$$ is at $$-1\,\text{m}$$ (1 m below sea-level).
• Point $$F$$ is at $$-3\,\text{m}$$.
• Point $$G$$ is at $$-6\,\text{m}$$.

Step 2 – Arrange the integers from biggest to smallest (decreasing order).

Since

the points in decreasing order of height are

Step 3 – Arrange the integers from smallest to biggest (increasing order).

Because

the points in increasing order of height are

Thus both required orders have been written by simply comparing the heights as integers.

We take mean sea level as the reference height $$0$$.

Any point lying above this level has a positive altitude. The point whose altitude is greatest will therefore be the highest point above sea level.

From geographical surveys it is known that the summit of Mount Everest, situated in the Himalayas (on the Nepal–China border), has the greatest positive altitude.

The measured height of Mount Everest is

Hence, the highest point on Earth above sea level is the peak of Mount Everest at about $$8848\,\text{m}$$.

Step 1 : Fixing the reference level
In geography we measure heights "above" or "below" the mean sea level (M.S.L.).
We agree to call the sea level the zero level, that is $$0\;\text{m}$$.

Step 2 : Understanding what “lowest point” means
A point that is below the sea level will have a negative height. The farther it is below the sea level, the smaller (more negative) the number will be.

Step 3 : Identifying the lowest point known
From measurements made by oceanographers, the deepest place found so far on the Earth is a part of the Mariana Trench in the Pacific Ocean. The particular spot is called the Challenger Deep.

Step 4 : Stating its depth as a signed number
The depth measured there is approximately
$$11\,034\;\text{m}$$ below the sea level.
Because it is below (on the other side of zero), we put a minus sign in front of the number:

Thus, in integer language, the height of the Challenger Deep with respect to the mean sea level is $$-11\,034\;\text{m}$$.

Result
The lowest point with respect to sea level, whether on land or on the ocean floor, is the Challenger Deep in the Mariana Trench, and its height (depth) is $$-11\,034\;\text{m}$$.

Step 1 : Collect some real winter temperature data

All these readings are negative numbers because the mercury column falls below the zero mark of the Celsius scale.

Step 2 : What is common among these places?

• They all lie in or very close to the Himalayan mountain ranges.
• Their altitudes are high. For example:
    Dras ≈ $$3300\,\text{m}$$, Leh ≈ $$3500\,\text{m}$$, Gulmarg ≈ $$2650\,\text{m}$$, Shimla ≈ $$2200\,\text{m}$$.
• Most of them are situated in the northern part of India (higher latitude).

Step 3 : Why does it become so cold there?

1. Effect of Altitude
   When we climb up a mountain, the air becomes thinner and cannot hold much heat. On an average, for every rise of $$1000\,\text{m}$$, the temperature falls by about $$6.5^{\circ}\mathrm{C}$$ (this is called the normal lapse rate). For example, if the temperature at sea level is $$15^{\circ}\mathrm{C}$$, then at Leh (≈ $$3500\,\text{m}$$) it can be estimated as
   \[15^{\circ}\mathrm{C}-3.5\times6.5^{\circ}\mathrm{C}\;\approx\;-8^{\circ}\mathrm{C}\] So high altitude alone can easily push the reading below zero.
2. Effect of Latitude
   Northern India is farther from the equator. Sun-rays reach these places at a more slanting angle, so the ground receives less heat.
3. Snow cover and clear skies
   Snow reflects (bounces back) most of the sunlight. During winter nights, clear skies allow rapid loss of heat from the ground, making the air even colder.

Step 4 : Why do most other places in India not cross the zero line?

• They are at low altitudes (near sea level) where the thick blanket of air stores and circulates heat.
• Many of them are near warm seas (the Arabian Sea, Bay of Bengal, Indian Ocean) whose water retains heat and keeps the adjoining land warm.
• Being closer to the equator, they get more direct sunlight.

Hence, only the high-mountain, northern-latitude stations listed above frequently record negative (below $$0^{\circ}\mathrm{C}$$) temperatures, while the rest of India generally stays on the positive side of zero.

Step 1 – Arrange the four times of the day in the order in which the Sun warms Leh.

• Early night — 11 p.m.
• Late night — 02 a.m.
• Late morning — 11 a.m.
• Early afternoon — 02 p.m.

Because the Sun is absent at night and highest in the sky in the afternoon, temperatures normally follow the opposite order (from lowest to highest):

coldest at 02 a.m. <  11 p.m. <  11 a.m. <  warmest at 02 p.m.

Step 2 – Arrange the four temperature readings from the coldest to the warmest.

$$-4^{\circ}\mathrm{C} \lt -2^{\circ}\mathrm{C} \lt 8^{\circ}\mathrm{C} \lt 14^{\circ}\mathrm{C}$$

Step 3 – Match the smallest temperature with the coldest time, the next with the next colder time, and so on.

The temperatures are now matched logically with the times of day.

Given grid

The middle (blank) square is not part of the border. We therefore list only the eight border numbers:

$$5,\; -3,\; -5,\; 0,\; -5,\; -8,\; -2,\; 7$$

Step 1 – Add the positive numbers

• $$5 + 7 = 12$$

Step 2 – Add the negative numbers

• $$-3 + (-5) = -8$$
• $$-8 + (-5) = -13$$
• $$-13 + (-8) = -21$$
• $$-21 + (-2) = -23$$

So, the sum of all negative numbers is $$-23$$.

Step 3 – Combine the two sums (0 adds nothing)

$$12 + (-23) + 0 = 12 - 23 = -11$$

Border sum

Idea behind the exercise
The eight cells that lie on the outside of the 3 × 3 board are called the border. In each of the three boards we already know some of the border numbers. We have to fill the remaining empty cells so that

$$\text{(sum of all 8 border numbers)} = \text{required border sum}.$$

There are many possible correct answers; one systematic way is:

1. add the border numbers that have already been supplied;
2. find the extra total still needed;
3. choose any convenient integers for the empty border cells whose sum equals that extra total;
4. put any suitable number (for example $$0$$) in the centre, because the centre does not affect the border sum.

Grid 1  (required border sum = $$+4$$)

Known border numbers: $$-10,\; -5,\; 9.$$

Their total is $$-10+(-5)+9=-6.$$

To reach the border sum $$+4$$ we still need $$+10$$ because

$$-6+10=4.$$

Fill the five empty border cells with any integers that add up to $$10$$, for example

$$3+4+1+1+1 = 10.$$

Border check:

$$-10+3+4+1+(-5)+9+1+1 = 4\;\checkmark$$

Grid 2  (required border sum = $$-2$$)

Known border numbers: $$6,\; 8,\; -5,\; -2.$$

Their total is $$6+8+(-5)+(-2)=7.$$

We therefore need an extra $$-9$$ because

$$7+(-9)=-2.$$

Choose four numbers that add to $$-9$$, for example

$$-4 + (-2) + (-2) + (-1) = -9.$$

Border check:

$$6+8+(-4)+(-2)+(-5)+(-2)+(-2)+(-1)=-2\;\checkmark$$

Grid 3  (required border sum = $$-4$$)

Known border numbers: $$7,\; -5.$$

Their total is $$7+(-5)=2.$$

We now need $$-6$$ more because

$$2+(-6)=-4.$$

There are six empty border cells, so the simplest choice is to put $$-1$$ in each of them:

Border check:

$$7+(-1)+(-1)+(-1)+(-5)+(-1)+(-1)+(-1)=-4\;\checkmark$$

All three grids now have the required border sums.

Step 1 – List the integers to be used
The grid has nine cells, so we must write the nine consecutive integers

$$-4,-3,-2,-1,0,1,2,3,4.$$

Step 2 – Find their total

\[(-4)+(-3)+(-2)+(-1)+0+1+2+3+4 = 0\]

Step 3 – Relate the border sum to the centre
If the number written in the middle square is $$x$$ and the required border sum is $$S_b$$, then

$$S_b + x = 0$$

(because every one of the nine numbers has been counted once).

Step 4 – Fix the centre for border sum $$-4$$
We want $$S_b=-4$$, so

$$-4 + x = 0 \;\Longrightarrow\; x = 4.$$

Therefore the centre square must contain 4. The other eight integers automatically add up to $$-4$$, so any ordering of them around the outside will work.

Step 5 – Two different correct fillings

Step 6 – Verification

Way 1 border sum:
$$(-1)+0+3+(-3)+(-4)+1+(-2)+2 = -4.$$

Way 2 border sum:
$$(-4)+(-3)+(-2)+0+1+2+3+(-1) = -4.$$

Since both arrangements give the required border sum, the problem is solved (and many other arrangements are also possible – the only compulsory entry is $$4$$ in the centre).

Step 1 – Recall the rule for filling the grids
In every grid the instruction was: “write integers so that the total of every row and of every column is zero.”
This gives one condition for each row and one condition for each column.

Step 2 – Count how many conditions we really get
• If the grid has m rows and n columns, we seem to have $$m+n$$ equations.
• But the grand total of the whole grid is the sum of all row–totals and also the sum of all column-totals. Therefore one equation is repeated.
So only $$m+n-1$$ equations are different from each other.

Step 3 – Count how many blank boxes there are
The grid itself has $$m\times n$$ unknown numbers (one in each box).

Step 4 – Compare unknowns with independent equations
The number of “free choices” that remain is

\[\text{free choices}=mn-(m+n-1)=(m-1)(n-1).\]

Step 5 – Decide when the grid has more than one solution
• If $(m-1)(n-1)=0$ (that happens only when m=1 or n=1), there are no free choices; the filling is unique.
• If $(m-1)(n-1)>0$ (that is, both m and n are at least 2) there is at least one free choice, hence many different fillings are possible.

Step 6 – State the result in simple words
Any grid that has more than one row and more than one column (2×2, 2×3, 3×3, …) can be filled in several ways. The reason is that after we meet the row–sum and column–sum rules there are still some boxes left which we can choose freely, and changing those numbers changes the whole filling without breaking the rules.

Example: 2 × 2 grid
Let the top-left box be $$k$$ (we can pick any integer).
• Top row must add to 0 → top-right becomes $$-k$$.
• Left column must add to 0 → bottom-left becomes $$-k$$.
• Finally bottom row (or right column) forces bottom-right to be $$k$$.
Because $$k$$ can be any integer, there are infinitely many different 2 × 2 fillings. The same freedom, only larger, exists for every bigger grid.

What is a “border integer square”?
Take an ordinary  3 × 3  square grid. The eight outer cells (shown below in grey) are called the border. A border integer square puzzle asks you to place integers in these eight cells so that each side of the square satisfies some given condition, usually a fixed sum.

Below you learn, step-by-step, how to design such a puzzle and, at the end, you get a ready-made challenge for your classmates (with the complete solution shown after the explanation).

Step 1 – Name the border cells

Give letters to the eight border cells so that you can write neat algebra sentences.

Step 2 – Choose the side-sums you want

• Top side (a + b + c)   → choose $$6$$
• Bottom side (g + f + e) → choose $$-3$$
• Left side (a + h + g)  → choose $$4$$
• Right side (c + d + e) → choose $$-2$$

(Any other four sums would also work, but integers that are not too large keep the puzzle friendly!)

Step 3 – Pick a few border numbers first

A quick way to keep every number small is to finish one whole side at once. Let us decide the whole top side right now:

• Choose $$a = 5$$, $$b = -1$$, $$c = 2$$
  (Check:  5 + (–1) + 2 = 6 — works!)

Step 4 – Use the chosen sums as equations

(1) Left column: $$a + h + g = 4$$ gives
$$5 + h + g = 4 \[3pt] \Rightarrow\; h + g = -1$$

Choose an easy pair for $h$ and $g$ that adds to –1; for instance $$h = 2,\; g = -3$$.

(2) Bottom row: $$g + f + e = -3$$ gives
$$-3 + f + e = -3 \[3pt] \Rightarrow\; f + e = 0$$

(3) Right column: $$c + d + e = -2$$ gives
$$2 + d + e = -2 \[3pt] \Rightarrow\; d + e = -4$$

So now we have a mini-system

Let the middle bottom cell be something nice, say $$f = 2$$. Then

(4) From $$f + e = 0 \;\Rightarrow\; 2 + e = 0 \;\Rightarrow\; e = -2$$

(5) From $$d + e = -4 \;\Rightarrow\; d + (-2) = -4 \;\Rightarrow\; d = -2$$

Step 5 – Write down the full, finished square

Check every side one last time:

• Top:  5 + (–1) + 2 = 6   ✓
• Bottom:  –3 + 2 + –2 = –3   ✓
• Left:  5 + 2 + –3 = 4   ✓
• Right:  2 + –2 + –2 = –2   ✓

How to turn this into a classroom puzzle

Erase four or five of those entries but keep all the side-sums visible. Your classmates now have to recover every blank using only integer arithmetic.

Here is a ready-made version you can photocopy or draw on the board (the shaded cells show the clues you leave; the centre is always blank):

Side-sums:
Top = 6,  Bottom = –3,  Left = 4,  Right = –2

When they are done, the only solution they can possibly reach is the complete square shown above. Have fun watching them race to finish it!

This activity is meant to check, by direct calculation, that the total you obtain depends on the particular integers you pick. Below we repeat the experiment three times, each time choosing different sets of six integers (three positive and three negative) and adding them carefully, step by step.

1. First attempt

   • Chosen positives: $$+7$$, $$+4$$, $$+3$$
   • Chosen negatives: $$-5$$, $$-2$$, $$-6$$

   Adding the positives first: $$7+4+3 = 14$$.

   Adding the negatives: $$(-5)+(-2)+(-6) = -(5+2+6) = -13$$.

   Now add the two results: $$14 + (-13) = 14-13 = 1$$.

   Sum obtained = $$1$$.

2. Second attempt

   • Chosen positives: $$+9$$, $$+6$$, $$+2$$
   • Chosen negatives: $$-4$$, $$-8$$, $$-5$$

   Positives: $$9+6+2 = 17$$.

   Negatives: $$(-4)+(-8)+(-5) = -(4+8+5) = -17$$.

   Add the two parts: $$17 + (-17) = 0$$.

   Sum obtained = $$0$$.

3. Third attempt

   • Chosen positives: $$+12$$, $$+1$$, $$+3$$
   • Chosen negatives: $$-10$$, $$-2$$, $$-7$$

   Positives: $$12+1+3 = 16$$.

   Negatives: $$(-10)+(-2)+(-7) = -(10+2+7) = -19$$.

   Add: $$16 + (-19) = 16-19 = -3$$.

   Sum obtained = $$-3$$.

What do we observe?

• First attempt gave $$1$$.
• Second attempt gave $$0$$.
• Third attempt gave $$-3$$.

Each time the sum changed because we selected a different set of integers. Therefore, the total is not fixed; it entirely depends on the particular positive and negative numbers you choose. Trying a few more times will keep giving different answers unless, by coincidence, the positives and negatives balance out in exactly the same way.

The rule of the game
Select exactly one number from every row in such a way that no two selected numbers come from the same column. Add the four chosen numbers. (In other words, you use each column once and each row once.)

I. Working with Grid 1

Two different trials are enough to convince ourselves that the total does not change.

Trial 1 (columns 1, 3, 2, 4):
$$7 + 4 + (-8) + (-11) = -8$$

Trial 2 (columns 4, 2, 1, 3):
$$16 + 1 + (-11) + (-14) = -8$$

Both give the same sum, $$-8$$. To see why this always happens, write every entry of the grid by means of its row–column position.

The column number $$c$$ can be $$0,1,2,3$$. Because we must use every column once, the four values of $$c$$ are exactly $$0,1,2,3$$. Hence

So Grid 1 always gives $$-8$$.

II. Working with Grid 2

Again take two sample trials.

Trial 1 (columns 1, 2, 3, 4):
$$-11 + (-6) + (-1) + 4 = -14$$

Trial 2 (columns 4, 1, 2, 3):
$$-8 + (-7) + (-2) + 3 = -14$$

Both totals are $$-14$$. The algebra is similar. Any entry can be written as $$-11 + 4r + c$$ where the row number $$r=0,1,2,3$$ and the column number $$c=0,1,2,3$$. Using every row and every column once gives

Therefore Grid 2 always gives $$-14$$.

Result
Whatever valid set of four numbers you pick,

Suppose we want three integers whose total is $$-8$$.

Let us choose the following numbers step by step.

1. First number: take $$a = -3$$.

2. Second number: take $$b = -4$$.

3. Now add these two: $$a + b = -3 + (-4) = -7$$.

   We still need $$-1$$ more (because $$-7 + ? = -8$$).
   The required extra amount is therefore $$-1$$.

4. Third number: take $$c = -1$$.

Check the sum of the three chosen numbers:

$$a + b + c = -3 + (-4) + (-1)$$

First add the first two: $$-3 + (-4) = -7$$.

Then add the third: $$-7 + (-1) = -8$$.

Thus the three numbers $$-3, -4, -1$$ satisfy the requirement.

Step 1 : List all numbers on one die

The faces of each die show these six numbers:

$$-5,\;-3,\;-1,\;2,\;4,\;6$$

So, while throwing two dice we may add any one of the six numbers on the first die to any one of the six numbers on the second die.

Step 2 : Make a table of every possible sum

Step 3 : Collect all different sums that really appear

Reading the table we get the following distinct sums:

$$-10,\;-8,\;-6,\;-4,\;-3,\;-2,\;-1,\;1,\;3,\;4,\;5,\;6,\;8,\;10,\;12$$

Step 4 : Write every whole number from $$-10$$ to $$12$$

$$-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12$$

Step 5 : Cross out the numbers that do occur (they are in Step 3).

The numbers that are not crossed out are impossible to obtain.

Missing numbers

$$-9,\;-7,\;-5,\;0,\;2,\;7,\;9,\;11$$

Hence these eight numbers cannot be obtained by adding the numbers on the two special dice.

Before working out each part, recall two key facts about integers:

• Every integer $$a$$ has an additive inverse (or opposite) $$-a$$ such that $$a + (-a) = 0$$.
• Subtraction rule: To subtract an integer, add its additive inverse.
    That is, $$a - b = a + (-b)$$.

Apply this rule carefully to every expression.

Thus each expression is evaluated, respecting the subtraction rule for integers.

We have six cards with the integers
$$+1,\; +7,\; +18,\; -5,\; -2,\; -9.$$

The task is to join some of them with addition and/or subtraction so that the value of the whole expression is as close as possible to $$-30$$.

Because we are allowed to use any card exactly once and may either add it or subtract it, each card can contribute one of two possible numbers:

• $$+1$$ or $$-1$$,
• $$+7$$ or $$-7$$,
• $$+18$$ or $$-18$$,
• $$-5$$ or $$+5$$,
• $$-2$$ or $$+2$$,
• $$-9$$ or $$+9$$.

Let us try to make the total exactly equal to $$-30$$ (that would clearly be the closest possible).

Notice the following combination:

• take $$-18$$ (that is, subtract the card $$+18$$),
• add $$-9$$ (simply write the card $$-9$$),
• add $$-5$$ (write the card $$-5$$),
• subtract $$-2$$ (that is, write $$-(-2) = +2$$).

The corresponding expression is

$$(-5) + (-9) - (+18) - (-2).$$

Now evaluate step by step.

1. First two terms:
   $$(-5) + (-9) = -(5+9) = -14.$$
2. Subtract the next card:
   $$-14 - (+18) = -14 - 18 = -(14+18) = -32.$$
3. Finally, subtract $$-2$$ (which is the same as adding $$+2$$):
   $$-32 - (-2) = -32 + 2 = -30.$$

Thus the value of the expression is exactly $$-30$$, which is as close as one can get to the required number.

One possible answer
$$\boxed{\;(-5) + (-9) - (+18) - (-2) = -30\;}. $$

(Other correct expressions are also possible; any expression that evaluates to $$-30$$ or as near as possible is acceptable.)

Step 1 · Understand the pattern
One complete block of the pattern contains five tokens:

$$+,\;+,\;+,\;-,\;-$$

In every such block:
• number of “+” tokens  = $$3$$
• number of “−” tokens  = $$2$$

Step 2 · Convert tokens into numbers
For integers, we can think of

• each “+” token as $$+1$$,
• each “−” token as $$-1$$.

So the value of one complete block is

$$3\times(+1)+2\times(-1)=3-2=1$$

Step 3 · Count how many complete blocks fit into 100 tokens
Each block has $$5$$ tokens.
Total tokens given  = $$100$$

Number of complete blocks:

$$\dfrac{100}{5}=20$$

Step 4 · Find the total value
Each block is worth $$1$$, and there are $$20$$ such blocks. Therefore

\[20\times1=20\]

Step 5 · Conclusion
The value of the entire string of 100 tokens is $$20$$.

Pre-reading  Brahmagupta (7^th century) called positive numbers fortunes and negative numbers debts. In NCERT Class 6 Chapter 10 we look at the same ideas through “Bela’s Building of Fun”. In that building

• Ground floor = floor 0
• Floors above ground are numbered $$+1,+2,+3,\ldots$$ (fortunes)
• Basements below ground are numbered $$-1,-2,-3,\ldots$$ (debts)

Walking up one floor means “add $$+1$$”; walking down one floor means “add $$-1$$”. Exactly the same moves appear on a horizontal number line: rightward is positive, leftward is negative.

Brahmagupta’s rules  ↔  Bela’s Building / Number line

Key take-away

• Positive numbers behave like floors above ground or right-hand walks on the number-line.
• Negative numbers behave like basements or left-hand walks.
• Brahmagupta’s 1 300-year-old rules are nothing but “go up, go down” statements, exactly what we practise on Bela’s Building and on a number line.

Rule 1 : Adding two positive integers

Take any two positive numbers, add them exactly the way you add whole numbers, the result remains positive.

• Example  1:  $$8+3=11$$
• Example  2:  $$15+27=42$$

Rule 2 : Adding two negative integers

Add their absolute (numerical) values and put a minus sign in front of the answer.

• Example  1:  $$(-6)+(-4)=-(6+4)=-10$$
• Example  2:  $$(-13)+(-5)=-(13+5)=-18$$

Rule 3 : Adding a positive and a negative integer

Subtract the smaller absolute value from the bigger one; the sign of the bigger absolute value stays.

• Example  1:  $$9+(-5)=9-5=4$$ (positive because 9>5)
• Example  2:  $$(-7)+3=-(7-3)=-4$$ (negative because 7>3)

Rule 4 : Subtracting an integer

To subtract an integer, add its additive inverse (change its sign).

• Example  1:  $$5-(-2)=5+2=7$$
• Example  2:  $$(-4)-6=(-4)+(-6)=-10$$