Let N denote the set of all natural numbers. Define two binary relations on N as $$R_1 = \{(x, y) \in N \times N : 2x + y = 10\}$$ and $$R_2 = \{(x, y) \in N \times N : x + 2y = 10\}$$. Then:

We have two relations on the set of natural numbers $$N$$.

The first relation is defined by $$R_1=\{(x,y)\in N\times N:2x+y=10\}$$. To list all ordered pairs of natural numbers that satisfy $$2x+y=10$$ we solve for every admissible integer value of $$x$$.

Let $$x=1\;(\text{natural}),\;2\cdot1+y=10\Rightarrow y=10-2=8.$$ Let $$x=2,\;2\cdot2+y=10\Rightarrow y=10-4=6.$$ Let $$x=3,\;2\cdot3+y=10\Rightarrow y=10-6=4.$$ Let $$x=4,\;2\cdot4+y=10\Rightarrow y=10-8=2.$$ Let $$x=5,\;2\cdot5+y=10\Rightarrow y=10-10=0,$$ which is not a positive natural number, so we stop here. Thus $$R_1=\{(1,8),\,(2,6),\,(3,4),\,(4,2)\}.$

The range of a relation is the set of all second components. Hence $$\text{Range}(R_1)=\{8,6,4,2\}=\{2,4,6,8\}.$$ Option C claims the range is $$\{2,4,8\},$$ which omits $$6,$$ so that statement is false.

Now consider symmetry for $$R_1$$. A relation is symmetric if $$\forall\,(x,y)\in R,\;(y,x)\in R.$$ Because $$(1,8)\in R_1$$ yet $$(8,1)\notin R_1$$ (since $$2\cdot8+1=17\neq10$$), $$R_1$$ is not symmetric. To test transitivity we recall: a relation is transitive if $$(x,y)\in R\;\text{and}\;(y,z)\in R\implies(x,z)\in R.$$ Take the pairs $$(4,2)\in R_1$$ and $$(2,6)\in R_1.$$ Their “middle” element is $$2,$$ so we must check whether $$(4,6)\in R_1.$$ But $$2\cdot4+6=14\neq10,$$ so $$(4,6)\notin R_1.$$ Hence $$R_1$$ is not transitive either.

The second relation is $$R_2=\{(x,y)\in N\times N:x+2y=10\}.$

Again we enumerate:

Let $$y=1,\;x+2\cdot1=10\Rightarrow x=10-2=8.$$ Let $$y=2,\;x+4=10\Rightarrow x=6.$$ Let $$y=3,\;x+6=10\Rightarrow x=4.$$ Let $$y=4,\;x+8=10\Rightarrow x=2.$$ Let $$y=5,\;x+10=10\Rightarrow x=0,$$ not a positive natural number, so we stop. Thus $$R_2=\{(8,1),\,(6,2),\,(4,3),\,(2,4)\}.$$

The range is the set of all second components, so $$\text{Range}(R_2)=\{1,2,3,4\}.$$ This matches exactly what Option B states.

For symmetry of $$R_2$$, observe $$(8,1)\in R_2$$ but $$(1,8)\notin R_2$$ because $$1+2\cdot8=17\neq10.$$ Hence $$R_2$$ is not symmetric.

For transitivity of $$R_2$$ we again use the definition. Take $$(6,2)\in R_2$$ and $$(2,4)\in R_2.$$ Because they chain through the element $$2,$$ we test $$(6,4).$$ Compute $$6+2\cdot4=14\neq10,$$ so $$(6,4)\notin R_2.$$ Thus $$R_2$$ is not transitive.

Summarising our findings:

• Neither relation is symmetric, so Option D is false. • Neither relation is transitive, so Option A is false. • The range of $$R_1$$ is $$\{2,4,6,8\},$$ not $$\{2,4,8\},$$ so Option C is false. • The range of $$R_2$$ is indeed $$\{1,2,3,4\},$$ so Option B is correct.

Hence, the correct answer is Option B.