Question

Prove that root 2 is irrational and hence prove
2+5root2 is irrational​

Answer 1

In mathematics, the irrational numbers are all the real numbers which are not rational numbers. That is, irrational numbers cannot be expressed as the ratio of two integers. ... For example, the decimal representation of π starts with 3.14159, but no finite number of digits can represent π exactly, nor does it repeat.

Answer 2

To ProvE :-

• $\sf \sqrt2 \ is \ a \ Irrational \ number .$
• Then prove that 2 + 5√2 is also Irrational.

ProoF :-

A number √2 is given and we need to prove that its Irrational . So on the contrary let us assume that √2 is a Rational Number . So it can be expressed in the form of p/q where p and q are integers and q ≠ 0 . Therefore ,

$\sf \to \sqrt{2}=\dfrac{p}{q}$

Also here p and q are co-primes. That is HCF of p and q is 1 .

$\qquad\qquad\tiny{\dag\red{\sf Squaring\ both \ sides , we have :- }}\\\\\sf:\implies\bigg( \dfrac{p}{q}\bigg)^2= (\sqrt2)^2\\\\\sf:\implies \dfrac{p^2}{q^2}=2 \\\\\sf:\implies 2p^2 = q^2$

This implies that 2 is a factor of q² . Subsequently by Fundamental Theorem of Arthemetic , we can say that 2 is also a factor of q . Let ,

$\sf:\implies \pink{ q = 2k }$

Now substituting this value in the initial equation we have ,

$\sf:\implies 2p^2=(2k)^2 \\\\\sf:\implies 2p^2 = 4k^2 \\\\\sf:\implies p^2 = 2k^2$

Again this implies that 2 is a factor of p² . Subsequently by Fundamental Theorem of Arthemetic , we can say that 2 is also a factor of p .

Therefore this contradicts our assumption that p and q are co-primes . Therefore our assumption was wrong . √2 is not a Rational Number . Therefore it is a Irrational Number.

$\rule{200}2$

Also we know that multiplication of Rational number and Irrational Number is Irrational. And sum of Rational number and Irrational Number is Irrational . Therefore 2 + 5√2 is Irrational.

$\boxed{\pink{\frak{ Hence \ Proved !! }}}$