Question

Show that $\sqrt{5}$ is an irrational number.

Answer 1

We have to show that $\sqrt{5}$ is irrational.

We will prove this via the method of contradiction.

So let's assume $\sqrt{5}$ is rational.

Hence, we can write $\sqrt{5}$ in the form $\frac{a}{b},$ where $a$ and $b$ are co-prime numbers such that $a,b,\in R$ and $b\ne 0.$

$\therefore \sqrt{5}=\frac{a}{b}$

squaring both sides we have

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

$\Rightarrow 5b^2=a^2$

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

Hence, 5 divides $a^2$

Now, a theorem tells that if 'P' is a prime number and P divides $a^2$ then P should divide 'a', where $a$ is a positive number.

Hence, 5 divides a $......\left(1\right)$

$\therefore$ we can say that $\frac{a}{5}=c$

we already know that

$\Rightarrow 5b^2=a^2$

From $\left(2\right),$ we know $a=5c$ substituting that in the above equation we get,

$\Rightarrow 5b^2=25c^2$

$\Rightarrow b^2=5c^2$

$\Rightarrow \frac{b^2}{5}=c^2$

Hence, 5 divides $b^2.$ And by the above mentioned theorem we can say that $5$ divides $b$ as well.

hence, $5$ divides $b$ $.........\left(3\right)$

So from $\left(2\right)$ and $\left(3\right)$ we can see that both a and b have a common factor $5.$ Therefore $a\&b$ are no co-prime. Hence our assumption is wrong. $\therefore$ by contradiction $\sqrt{5}$ is irrational.

Hence, solved.