Tag: Class 10th Maths Chapter 1

Exercise 1.2 — Solutions

1. Prove that $\sqrt{5}$ is irrational.

Proof (by contradiction).
Assume $\sqrt{5}$ is rational. Then there exist integers $a,b$ with $b\mathrm{\ne }0$ and $\gcd (a,b)=\mathrm{1,}$

such that

$$\sqrt{5}=\frac{a}{b}\mathrm{.}$$

Squaring both sides,

$$5=\frac{a^2}{b^2}\Rightarrow a^2=5b^2\mathrm{.}$$

So $a^2$ is divisible by 5, hence 5 divides $a$. Put $a=5c$for some integer $c$. Substituting,

$$(5c{)}^2=5b^2\Rightarrow 25c^2=5b^2\Rightarrow b^2=5c^2\mathrm{.}$$

Thus $b^2$ (and so $b$) is divisible by 5, which contradicts $\gcd (a,b)=1$. Therefore $\sqrt{5}$ is irrational. ∎

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2. Prove that $\sqrt{3}+\sqrt{5}$ is irrational.

Proof (by contradiction).
Assume $\sqrt{3}+\sqrt{5}$ is rational. Let

$$\sqrt{3}+\sqrt{5}=r$$

for some rational $r$. Then $\sqrt{3}=r-\sqrt{5}$. Squaring,

$$3=(r-\sqrt{5}{)}^2=r^2-2r\sqrt{5}+5.$$

Rearrange to isolate $\sqrt{5}$:

$$2r\sqrt{5}=r^2+5-3=r^2+2\Rightarrow \sqrt{5}=\frac{r^2+2}{2r}\mathrm{.}$$

The right-hand side is rational, so $\sqrt{5}$ would be rational — contradicting result 1. Hence $\sqrt{3}+\sqrt{5}$ is irrational. ∎

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3. Prove that the following are irrational:

(i) ${\displaystyle \frac{1}{\sqrt{2}}}$

If ${\displaystyle \frac{1}{\sqrt{2}}}$ were rational then its reciprocal $\sqrt{2}$would be rational (reciprocal of a nonzero rational is rational). But $\sqrt{2}$ is known to be irrational. Contradiction. Therefore ${\displaystyle \frac{1}{\sqrt{2}}}$ is irrational. ∎

(ii) $\sqrt{7}+\sqrt{5}$

Assume $\sqrt{7}+\sqrt{5}=\mathrm{r }$is rational. Then $\sqrt{7}=r-\sqrt{5}$. Squaring,

$$7=r^2-2r\sqrt{5}+5\Rightarrow 2r\sqrt{5}=r^2-2.$$

So

$$\sqrt{5}=\frac{r^2-2}{2r},$$

which is rational — contradicting result 1. Hence $\sqrt{7}+\sqrt{5}$ is irrational.

(iii) $\sqrt{6}+\sqrt{2}$

Assume $\sqrt{6}+\sqrt{2}=r$ is rational. Square both sides:

$$(\sqrt{6}+\sqrt{2}{)}^2=r^2\Rightarrow 6+2+2\sqrt{12}=r^2\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$$

Thus

$$2\sqrt{12}=r^2-8\Rightarrow \sqrt{12}=\frac{r^2-8}{2}\mathrm{.}$$

The right-hand side is rational, so $\sqrt{12}$ would be rational. But $\sqrt{12}=2\sqrt{3}$, so this would make $\sqrt{3}$ rational — contradiction. Therefore $\sqrt{6}+\sqrt{2}$ is irrational.