Question

proved that 1/√2 is an irrational number​

Answer 1

Answer:

▶Here's your answer

⏩Let us assume that 1/√2 is rational so...

$\frac{1}{ \sqrt{2} } = \frac{p}{q}$

⏩Where p and q are co prime and q id not equal to zero...

$\sqrt{2} q = p \\ \\ squaring \: both \: side \\ \\ { \sqrt{2q} }^{2} = {p}^{2} \\ \\ 2 {q}^{2} = {p}^{2}$

⏩Here 2/p ^2, 2/p then... p =2a

$2 {q}^{2} = ( {2a})^{2} \\ \\ 2 {q}^{2} = 4 {a}^{2} \\ \\ {q}^{2} = 2 {a}^{2}$

⏩2/q^2, 2/q

⏩There fore p and q has 2 as their common factors ....

⏩But this contradict the fact that p and q are co prime...

⏩ this contradiction has arising due to an incorrect assumption that 1 / root 2 is irrational..

⏩Therefore 1/√2 is irrational..

Hope it helps ✌

Answer 2

Answer :

We have to prove 1/√2 is an irrational number.

Let us assume the opposite,

i.e., 1/√2 is a rational number.

Hence 1/√2 can be written in the form a/b

where a and b are co-prime

Hence, 1/√2 = a/b