Question

√2 is Irrrational number proof that​

Answer 1

solution:-

here , it is Given √2 is irrational number.

Let √2 = p / q wher p,q are integers q ≠ 0

we also suppose that p / q is written in the simplest form

Now √2 = p / q⇒ 2 = p2 / q2 ⇒ 2q2 = p2

∴ 2q2 is divisible by 2

⇒ p2 is divisible by 2  

⇒ p is divisible by 2  

∴ let p = 2r

p2 = 4r2 ⇒ 2q2 = 4r2 ⇒ q2 = 2r2

∴ 2r2 is divisible by 2

∴ q2 is divisible by 2

∴ q is divisible by 2

∴p are q are divisible by 2 .

this contradicts our supposition that p/q is written in the simplest form

Hence, our supposition is wrong

∴ √2 is irrational number.

Answer 2

Answer:A proof that the square root of 2 is irrational

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.

Step-by-step explanation: