Question

Prove that the sum of a rational number and an irrational number is always irrational.

Answer 1

Proving that the sum of a rational number and an irrational number is always irrational.

Step-1: Given information:

Let $a$ be the rational number and $b$ be the irrational number.

To prove, $\mathrm{a}+\mathrm{b}$ is irrational.

To prove this by contradiction, assuming $\mathrm{a}+\mathrm{b}$ is rational.

By definition of rationals, we can write

$\mathrm{a}=\frac{\mathrm{p}}{\mathrm{q}},\mathrm{q}\ne 0$ and $\mathrm{a}+\mathrm{b}=\frac{\mathrm{m}}{\mathrm{n}},\mathrm{n}\ne 0$.

Step-2: Checking the truth of the given statement:

Substituting $\mathrm{a}=\frac{\mathrm{p}}{\mathrm{q}}$ in $\mathrm{a}+\mathrm{b}=\frac{\mathrm{m}}{\mathrm{n}}$

$\begin{array}{rcl}a+b& =& \left(a+b\right)\\ & \Rightarrow & \frac{\mathrm{p}}{\mathrm{q}}+\mathrm{b}=\frac{\mathrm{m}}{\mathrm{n}}\\ & \Rightarrow & \mathrm{b}=\frac{\mathrm{m}}{\mathrm{n}}-\frac{\mathrm{p}}{\mathrm{q}}\\ & \Rightarrow & \mathrm{b}=\frac{\mathrm{m}}{\mathrm{n}}+\left(-\frac{\mathrm{p}}{\mathrm{q}}\right)\end{array}$

Since rational numbers are closed under addition, $\mathrm{b}$ is a rational number.

This gives us the contraction.

Therefore, our assumption was wrong.

Thus, $\mathrm{a}+\mathrm{b}$ is irrational.

Hence, the given statement has been proven.