Question

prove that
$\sqrt{2} \: is \: irrational$
​

Answer 1

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.

• Notice that in order for a/b to be in simplest terms, both of aand 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. 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.

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.

If we substitute a = 2k into the original equation

2 = a2/b2,

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.

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 √2cannot be rational.

Hope it helps u....pls mark it as the brainliest ..

Answer 2

given :- √2 :- let us suppose that √2 is irrational.

➡ √2 = p/q (q ≠ 0, p and q are coprime)

➡ √2q = p

squaring both sides,

➡ (√2q)² = (p)²

➡ 2q² = p² ----------(i)

therefore 2 divides p ---------(ii)

➡ 2 × r = p

again squaring both sides,

➡ (2 × r)² = (p)²

➡ 4 × r² = p²

➡ 4 × r² = 2q² (from equation (i)

➡ 2 × r² = q²

2 divides q², therefore 2 divides q ---------(iii)

from (ii) and (iii), we can say that p and q have common factor 2 which contradicts our assumption that p and q are coprime.

hence, our assumption is wrong and it's proved √2 is irrational.