In proof of SQRT(2) is irrational, why can't a,b both be even

In summary, the classic proof of irrationality of SQRT(2) assumes that it can be represented by a rational number, a/b where a and b are integers. However, after a few mathematical steps, this leads to a contradiction where both a and b are even numbers. This is because the assumption of a and b being integers with no common factors results in a loop of reducing them to even numbers, which is impossible as the natural numbers have a minimum. Another proof involves reducing a and b to have no common factors and showing that this leads to a contradiction as well. Vi Hart's video provides an amusing take on Pythagoras and the proof of the irrationality of 2.
  • #1
Thecla
131
10
In the classic proof of irrationality of SQRT(2) we assume that it can be represented by a rational number,a/b where a, b are integers. This assumption after a few mathematical steps leads to a contradiction, namely that both a, b are even numbers.
Why is that a contradiction?
Well you can say that the rational fraction has to be in its lowest terms;therefore either a or b or both must be odd.
However, that wasn't in the assumption(lowest terms). The assumption was just two integers a,b.
Why can't they both be even?
 
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  • #2
In the correct version of the classic proof, it is assumed that a and b are not both even.
 
  • #3
And the reason you want to make the assumption that both are even right at the start is...
 
  • #4
The proof I use starts by assuming a and b are integers with no common factors i.e. a/b is in its lowest terms. Then you end up with a loop in which a and b always have a common factor of 2 - but this is impossible as the numerator and denominator would get smaller without limit. An impossibility for positive integers.
 
  • #5
Thecla said:
In the classic proof of irrationality of SQRT(2) we assume that it can be represented by a rational number,a/b where a, b are integers. This assumption after a few mathematical steps leads to a contradiction, namely that both a, b are even numbers.
Why is that a contradiction?
Well you can say that the rational fraction has to be in its lowest terms;therefore either a or b or both must be odd.
However, that wasn't in the assumption(lowest terms). The assumption was just two integers a,b.
Why can't they both be even?

If a and b are both even, then let ##a=2c## and ##b=2d##. Repeat the argument on c and d, so we conclude they too are even. Let ##c=2e## and ##d=2f##...which are also even. And so on. This argument should keep going forever.

But here's the thing. The natural numbers have a minimum (namely 1). So we can't keep halving forever. At some point this process must terminate. This contradicts our original argument.
 
  • #6
Thanks for the help
 
  • #7
Another proof:

In ##2 = (a/b)^2##, assume that ##a## and ##b## have been reduced to have no common factors.
Then ##2b^2=a^2##, so ##a## must be even (contain a factor 2).
Thus ##a=2c## for some ##c##
Thus ##2b^2=4c^2##
Thus ##b^2=2c^2##, so now ##b## must be even also.

This contradicts the assumption that ##a## and ##b## have no common factor, so no such ##a## and ##b## can exist.
 
  • #8
Just do this

Suppose $$\sqrt{2}=a/b$$
where
a=2^m c
b=2^n d
c and d are odd integers
then
$$
\sqrt{2}=a/b \\
\sqrt{2}b=a \\
2b^2=a^2 \\
2(2^n d)^2=(2^m c)^2 \\
2^{2n+1} d^2=2^{2m} c^2 \\
$$
thus
2n+1=2m
2n+1 is odd while 2m is even
an even number cannot equal an odd number
contradiction
 
Last edited:
  • #9
If you have never seen Vi Hart's videos, here is an amusing one on Pythagoras and the proof that 2 is irrational.

P.S. While you are there, check out her Mobius strip story of "Wind and Mr Ug"
 

1. Why can't both a and b be even in the proof of SQRT(2) is irrational?

The proof relies on the assumption that a and b are both integers and have no common factors. If both a and b are even, they would have a common factor of 2, which would contradict the initial assumption.

2. What if both a and b are odd in the proof of SQRT(2) is irrational?

If both a and b are odd, then they would have a common factor of 1, which does not contradict the initial assumption. However, this would result in an odd number when a^2 is divided by b^2, which would not equal 2. Therefore, the proof still holds.

3. Can one of the numbers, a or b, be even in the proof of SQRT(2) is irrational?

Yes, one of the numbers can be even. However, the other number must be odd in order for there to be no common factors between them. This is necessary for the proof to hold.

4. Why is it important that a and b have no common factors in the proof of SQRT(2) is irrational?

If a and b have common factors, then they can be simplified. This means that the square root of 2 would not be in its simplest form, which would contradict the definition of an irrational number.

5. Can this proof be applied to other square roots?

Yes, this proof can be applied to any square root of a non-perfect square (a number that is not a perfect square). The only difference would be using the square root of that number instead of 2 in the proof.

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