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Real Numbers — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsReal Numbers3 Marks
Question
Prove that for any prime positive integer $\text{p},\sqrt{\text{p}}$ is an irrational number.
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Answer

Given that p is a prime positive integer.
Let us assume p is rational number and there exist two positive integer a and b which are co-prime such that
$\sqrt{\text{p}}=\frac{\text{a}}{\text{b}}\ \dots(1)$
squaring both sides,
$\Rightarrow\ \text{p}=\frac{\text{a}^2}{\text{b}^2}$
$\Rightarrow\ \text{pb}^2=\text{a}^2\ \dots(2)$
$\Rightarrow\ \text{p|a}^2$
$\Rightarrow\ \text{p|a}\ \dots(3)$
Let a = pc some positive integer c.
put a = pc in equation (2)
$\Rightarrow\ \text{pb}^2=\text{p}^2\text{c}^2$
$\Rightarrow\ \text{b}^2=\text{pc}^2$
$\Rightarrow\ \text{p|b}^2$
$\Rightarrow\ \text{p|b}\ \dots(4)$
From equation (3) and (4), p is common factor of a and b.
This is the constraction of our assumption because a and b are co-prime (no common factor other than.)
Thus, $\sqrt{\text{p}}$ is an irrational numbers.

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