Why is square root of two an irrational number in Euclidian geometry?

Question

This is from by Philosopher relative who thought every engineer is a mathematician!!

Answer 1

I believe the question is in fact asking for a geometric proof of the irrationality of the square root of 2. There is one due to Apostol found below. But basically the idea of the proof is that assuming that sqrt(2) is rational, one can use Fermat's method of infinite descent by constructing triangles with sides of increasingly smaller integer length.

http://blog.plover.com/math/sqrt-2-new.html

Answer 2

This is true everywhere, not just in Euclidean geometry.
The following proof is from literature

http://en.wikipedia.org/wiki/Irrational_number

One proof of the irrationality of the square root of 2 is the following reductio ad absurdum. The proposition is proved by assuming the contrary and showing that doing so leads to a contradiction (hence the proposition must be true).

1. Assume that √2 is a rational number. This assumption implies that there exist integers m and n with n ≠ 0 such that m/n = √2.
2. Then √2 can also be written as an irreducible fraction m/n (the fraction is shortened as much as possible). This means that m and n are coprime integers, i.e., they have no common factor greater than 1.
3. From m/n = √2 it follows that m = n√2, and so m2 = (n√2)2 = 2n2.
4. So m2 is an even number, because it is equal to 2n2, which is even.
5. It follows that m itself is even (since only even numbers have even squares).
6. Because m is even, there exists an integer k satisfying m = 2k.
7. We may therefore substitute 2k for m in the last equation of (3), thereby obtaining the equation (2k)2 = 2n2, which is equivalent to 4k2 = 2n2 and may be simplified to 2k2 = n2.
8. Because 2k2 is even, it now follows that n2 is also even, which means that n is even (recall that only even numbers have even squares).
9. Then, by (5) and (8), m and n are both even, which contradicts the property stated in (2) that m/n is irreducible.

Since we have found a contradiction, the initial assumption (1) that √2 is a rational number is false; that is to say, √2 is irrational.

This proof can be generalized to show that any root of any natural number is either a natural number or irrational.

Answer 3

What does it have to do with Euclidian geometry?

Let's assume that there are two integers m,n such that

sqrt(2) = m/n and m/n is an irreducible vulgar fraction.

2 = m^2 / n^2

2n^2 = m^2

Thus m^2 is divisible by 2. Because 2 is a prime number
m is divisible by 2.

And thus, m^2=2n^2 is divisible by 4

If 2n^2 is divisible by 4 n^2 is divisible by 2.

Thus 2 is a common factor of m and n, which contradicts our assumption that m/n is irreducible.

Answer 4

1.4142135623730950488016887242097 is standing on the ledge of a 15 story building. It is not rational and we need somebody to talk it off the ledge. How about you? Tell it that it shouldn't feel so bad. Look at pi, it's way more irrational, but you don't see it wanting to take a flying leap. Good luck with that.

Answer 5

Because any number times itself will not equal two.

Answer 6

God has the answer. I'm pretty sure the answer is in Nature. The best laboratory for mathematical 'experiments'

Answer 7

So, you want to know, are you just incredulous that this relative asked this?

Answer 8

:-|

im glad i dun start school til September lol