Square root of two

1. Aug 30, 2014

johann1301

1. The problem statement, all variables and given/known data
In this task we will show that √2 is irrational. Assume that √2 = a/b where both a and b are natural numbers. Let a = p1p2p3...pn, and b = q1q2q3...qm be the prime factors

a) explain why 2q1q1q2q2q3q3...qmqm = p1p1p2p2p3p3...pnpn

√2=a/b

b√2=a

√2q1q2q3...qm=p1p2p3...pn

(√2(q1q2q3...qm))2=(p1p2p3...pn)2

2q1q1q2q2q3q3...qmqm = p1p1p2p2p3p3...pnpn

b) explain why there can't be the same amount of 2-factors on the left and the right

Im stuck on this one...

Is it because then a and b could be further shortened/divided?

2. Aug 30, 2014

pasmith

Is there some reason why you didn't write that as $$2q_1^2 \dots q_m^2 = p_1^2 \dots p_n^2?$$

The left hand side is even. Therefore the right hand side is also. But the right hand side is the square of an integer, $a$. Either $a$ is even or $a$ is odd. Only one of these is possible if $a^2$ must be even. What does that say about the minimum number of times $a^2$ can be divided by 2? What can you then say about the integer $b^2 = a^2/2$?

3. Aug 31, 2014

johann1301

No, the book wrote it that way. So i did to.

I don't know how to prove it, but i remember that if a2 is even, then a is even also(given that a2 is a square number which we know it is, because a was a whole number). So a is even and that would imply that i could write a=(2k) where k is a natural number 1, 2, 3, 4... If i square this i get that a2=(2k)2=4k2. I see from this that a2 could be divided by 2 two times and still be a natural number. If i know look at b2=a2/2 i can see that a2/2 can be written as 2k2.

We get that:
b2=2k2

b2/k2=2

b/k=√2

This last result goes against what what we assumed. That a and b had no common factors? (the task doesn't really assume this, i assume this)

(Or maybe we don't have to assume that a and b had no common factors. We can rather argue that since we originally had √2=a/b and got that this could be rewritten to √2=b/k, we could just repeat the process an infinite amount of times, and thus removing an infinite amount of 2-factors from the fraction. And this is absurd. So there can't be the same amount of 2-factors on the left and the right.)

Last edited: Aug 31, 2014
4. Aug 31, 2014

PeroK

You seem to be mixing up two different proofs. The simplest, which I think you've almost got above, is:

Assume a, b have no common factors, then show that a and b are both even. You don't need prime factorisations for this.

There is another proof using the unique prime factorisation of a and b. This can be used to show that if a and b have no common factors, then $a^2$ and $b^2$ have no common factors. Therefore, the square root of any whole number is either a whole number or irrational; it's never a/b with b ≠ 1.

5. Aug 31, 2014

johann1301

there is also a (last) task c)

Use the fundamental theorem of arithmetic to derive a contradiction
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Wouldn't i be doing task c if i do what you say?

I think I'm just supposed to - at least for now - explain why there can't be the same amount of 2-factors on the left and the right...

6. Aug 31, 2014

pasmith

Is the number of factors of 2 on the left even or odd?
Is the number of factors of 2 on the right even or odd?