Question

prove that √2 is irrational.​

Answer 1

Heya mate... here ya go...

Let's suppose √2 is a rational number. Then we can write it √2  = a/b where a, b are whole numbers, b not zero.

We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.

From the equality √2  = a/b it follows that 2 = a2/b2,  or  a2 = 2 · b2.  So the square of a is an even number since it is two times something.

From this we know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don't believe me!

Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

2 = (2k)2/b2

2 = 4k2/b2

2*b2 = 4k2

b2 = 2k2

This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!!

WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.

Have a good day ahead bro...

Answer 2

QuesTion

• Prove that √2 is irrational.

AnsweR

Step By Step Explanation

To Prove

• √2 is irrational

Solution

Let us assume to the contrary that √2 is rational.

So we can find co-prime integers a and b such that b ≠ 0 and HCF (a, b) = 1 .

$\frac{a}{b} = \sqrt{2}$

By squaring both sides,

$\implies \sqrt{ {2}} ^{2} = \: \frac{ {a}^{2} }{ {b}^{2} }$

$\implies 2 = \frac{{a}^{2} }{ {b}^{2} }$

$\implies 2 {b}^{2} \: = \: {a}^{2} \: ... \: \: (eq ^{n} 1)$

$\implies {a}^{2} \: is \: divisible \: by \: 2.$

$\therefore a \: is \: also \: divisible \: by \: 2$

Now, a = 2c , for any integer c.

By Putting The Value Of a in equation 1

$\implies \: 2 {b}^{2} \: = \: (2c) ^{2}$

$\implies \: 2 {b}^{2} \: = \: 4 {c}^{2}$

$\implies \: {b}^{2} \: = \: 2 {c}^{2}$

$\implies \: {b}^{2} \: is \: divisible \: by \: 2$

Therefore, b is also divisible by 2.

$\red\implies\red{a \: and \: b \: has \: 2 \: as \: a \: common \: fator}$

This contradicts the fact that a and b are co-primes.

This contradiction has arisen because of our wrong assumption , that √2 is rational.

$\therefore \: \sqrt{2 } \: \: is \: \: irrational$

Hence Proved .