Proving √2 Irrational - Explanation & Solution

[johann1301]

Homework Statement

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...p_n, and b = q₁q₂q₃...q_m be the prime factors

a) explain why 2q₁q₁q₂q₂q₃q₃...q_mq_m = p₁p₁p₂p₂p₃p₃...p_np_n

√2=a/b

b√2=a

√2q₁q₂q₃...q_m=p₁p₂p₃...p_n

(√2(q₁q₂q₃...q_m))²=(p₁p₂p₃...p_n)²

2q₁q₁q₂q₂q₃q₃...q_mq_m = p₁p₁p₂p₂p₃p₃...p_np_n

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?

 

[pasmith]

Homework Statement

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...p_n, and b = q₁q₂q₃...q_m be the prime factors

a) explain why 2q₁q₁q₂q₂q₃q₃...q_mq_m = p₁p₁p₂p₂p₃p₃...p_np_n

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

√2=a/b

b√2=a

√2q₁q₂q₃...q_m=p₁p₂p₃...p_n

(√2(q₁q₂q₃...q_m))²=(p₁p₂p₃...p_n)²

2q₁q₁q₂q₂q₃q₃...q_mq_m = p₁p₁p₂p₂p₃p₃...p_np_n

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?

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

 

[johann1301]

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

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

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

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.)

 

[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.

 

[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?

Assume a, b have no common factors, then show that a and b are both even.

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...

 

[pasmith]

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...

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?