Question

prove that √2 is irrational



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Answer 1

Answer:

Here is ur answer:

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

√2=a/b --- (a&b are any co-prime)

√2b=a ---squaring on both sides

2b^2=a^2---(a^2 is divisible by 2)

(a is also divisible by 2)

a=2c ----(C is any interger)

2b^2=2C

2b=c ------(c^2 is divisible by 2)

(c is also divisible by 2)

So, it is irrational no.

This contradiction have Arristion due to our incorrect assumptions

So, we can say that √2 is irrational no.

Answer 2

We should assume √2 is a sane number. At that point we can compose it √2 = a/b where a, b are entire numbers, b not zero.

We also expect to be that this a/b is rearranged to least terms, since that should clearly be possible with any division. Notice that all together for a/b to be in least difficult terms, both of an and b can't be even. One or both should be odd. Else, we could improve on a/b further.

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

From this we realize that an itself is additionally a significantly number. Why? Since it can't be odd; on the off chance that an itself was odd, a · an eventual odd as well. Odd number occasions odd number is consistently odd. Check it on the off chance that you don't trust me!

Alright, in the event that an itself is a significantly number, an is multiple times some other entire number. In images, a = 2k where k is this other number. We don't have to understand what k is; it will not make any difference. Before long comes the inconsistency.

On the off chance that we substitute a = 2k into the first condition 2 = a2/b2, this is the thing that we get:

2 = (2k)2/b2

2 = 4k2/b2

2*b2 = 4k2

b2 = 2k2

This implies that b2 is even, from which follows again that b itself is even. Furthermore, that is a logical inconsistency!!!

Why would that be a logical inconsistency? Since we began the entire interaction expecting to be that a/b was improved to most reduced terms, and now incidentally, an and b both would be even. We finished at an inconsistency; along these lines our unique presumption (that √2 is reasonable) isn't right. Consequently √2 can't be objective.