Question

prove that the following are irrational
$1 \ \sqrt{2 \:}$
​

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Let us assume that 1√2 is a rational number.

So,

$\hookrightarrow\tt{1\sqrt{2}=\dfrac{p}{q}}$

$\hookrightarrow\tt{\sqrt{2}=\dfrac{p}{q}}$

Now, we can see that the left hand side is irrational where as the right hand side is rational.

So, our assumption is contradicted.

And hence, 1√2 is an irrational number.

$\rule{200}2$

We can solve such questions easily if we can prove for example √n as an irrational number.

So, again let us assume that √n is a rational number.

Hence,

$\hookrightarrow\tt{\sqrt{n}=\dfrac{P}{Q}}$

$\hookrightarrow\tt{\sqrt{n}\times Q= P}$

Squaring both the sides,

$\hookrightarrow\tt{{(\sqrt{n}\times Q)}^{2}={(P)}^{2}}$

$\hookrightarrow\tt{n\times{Q}^{2}={P}^{2}}$......(1)

$\hookrightarrow\tt{{Q}^{2}= \dfrac{{P}^{2}}{n}}$

Now, we can see that n divides P² so it also divides P.

We know that,

Any prime number if divides a², then it can also divide a, where a is an integer.

So,

$\hookrightarrow\tt{\dfrac{P}{n}=c}$

( For c as some integer )

$\hookrightarrow\tt{P= n\times c}$

Substituting it in (1),

$\hookrightarrow\tt{{(n\times c)}^{2}=n\times{Q}^{2}}$

$\hookrightarrow\tt{{n}^{2}\times{P}^{2}=n\times{Q}^{2}}$

$\hookrightarrow\tt{{Q}^{2}=\dfrac{{n}^{2}\times{p}^{2}}{n}}$

$\hookrightarrow\tt{{Q}^{2}=n\times{c}^{2}}$

$\hookrightarrow\tt{{c}^{2}=\dfrac{{Q}^{2}}{n}}$

Now, we see that n devides Q², and also divides Q.

( Any prime number if divides a², then it can also divide a, where a is an integer. )

So, n divides both P and Q.

So, our presumption is wrong.

Hence, √n is an irrational number.

$\rule{200}4$

Answer 2

$\huge \underline{ \rm \red{hello}}$

Question:-

Prove that following is irrational number.

$1\sqrt{2}$

Solution :-

To prove 1/√2 is irrational

Let us assume that √2 is irrational 

1/√2 = p/q (where p and q are co prime)

q/p = √2

q     = √2p

squaring both sides

q²   = 2p²                                                  .....................(1)

By theorem 

q is divisible by 2

∴ q = 2c ( where c is an integer)

 putting the value of q in equitation 1

2p² = q² = 2c² =4c²

p² =4c² /2 = 2c²

p²/2 = c² 

by theorem p is also divisible by 2

But p and q are coprime

This is a contradiction which has arisen due to our wrong assumption

∴1/√2 is irrational