Question

proove that root 7 is irrational​

Answer 1

First assume that √7 is rational.

So let √7 be written in the form of a/b, which is of the simplest form where a, b are co-prime integers .

$\frac{a}{b}=\sqrt{7}\\ \\ (\frac{a}{b})^2=(\sqrt{7})^2\\ \\ \frac{a^2}{b^2}=7 \\ \\ a^2=7b^2$

Here we get that a² is a multiple of 7.

So a will also be a multiple of 7.  [As 'a' is a positive integer, a² is a perfect square.]

So let a = 7m.

$a^2=7b^2 \\ \\ (7m)^2=7b^2 \\ \\ 49m^2=7b^2 \\ \\ 7m^2=b^2$

Here it seems that b² is also a multiple of 7.

So b is also a multiple of 7.

But this contradicts the earlier assumption that a, b are co-prime integers. Because now it seems that 7 is a common factor of a and b.

∴ √7 is irrational.

Thank you. Have a nice day. :-))

#adithyasajeevan