Question

proov that root 2 is not a rational number step by step ​

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, 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.

Follow me dear for your future doubts ❤️❤️

Answer 2

Answers

Let √2 be a rational number

Therefore, √2= p/q [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0

On squaring both sides, we get

p²= 2q² ...(1)

Clearly, 2 is a factor of 2q²

⇒ 2 is a factor of p² [since, 2q²=p²]

⇒ 2 is a factor of p

Let p =2 m for all m ( where m is a positive integer)

Squaring both sides, we get

p²= 4 m² ...(2)

From (1) and (2), we get

2q² = 4m² ⇒ q²= 2m²

Clearly, 2 is a factor of 2m²

⇒ 2 is a factor of q² [since, q² = 2m²]

⇒ 2 is a factor of q

Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1

Therefore, Our supposition is wrong

Hence √2 is not a rational number i.e., irrational number.

hope its help u.