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Proof that sq root of 2 is irrational

Hi. I don’t suppose you get too many retired accountants blogging here, but as I had nothing better to do I took Stephen Hawking’s “Brief History of Time” on holiday with me this year. Load of waffle (how come string theory is not disproved by the two-slit interference experiment?). Anyway, that’s not my question. As I am still no wiser as to why light travels at the same speed for every observer, I decided to look up some physics lectures on the internet, and came across the concept of irrational numbers, which I had totally and utterly forgotten about (not used by accountants – or so we claim. Imaginary numbers maybe, but not irrational ones!).

Apparently the Pythagoreans claim to have proved that the square root of two is irrational by examining the equation a^2 = 2*b^2. They observed that the square of any even number is even, and that of any odd number is odd, which I don’t have any difficulty with. However the proof requires reverse symmetry, as it claims that because a^2 is even then so must be a. How come? If you set b=3 then a^2 = 18. a^2 is certainly even, but the square root of 18 seems anything but either rational or even! Even if it is, it is no more obvious to me than the original conundrum, so we are just going round in circles!

What have I missed here?

Reply 1

lets_make_some_music

JohnP

Hi. I don’t suppose you get too many retired accountants blogging here, but as I had nothing better to do I took Stephen Hawking’s “Brief History of Time” on holiday with me this year. Load of waffle (how come string theory is not disproved by the two-slit interference experiment?). Anyway, that’s not my question. As I am still no wiser as to why light travels at the same speed for every observer, I decided to look up some physics lectures on the internet, and came across the concept of irrational numbers, which I had totally and utterly forgotten about (not used by accountants – or so we claim. Imaginary numbers maybe, but not irrational ones!).

Apparently the Pythagoreans claim to have proved that the square root of two is irrational by examining the equation a^2 = 2*b^2. They observed that the square of any even number is even, and that of any odd number is odd, which I don’t have any difficulty with. However the proof requires reverse symmetry, as it claims that because a^2 is even then so must be a. How come? If you set b=3 then a^2 = 18. a^2 is certainly even, but the square root of 18 seems anything but either rational or even! Even if it is, it is no more obvious to me than the original conundrum, so we are just going round in circles!

What have I missed here?

is it (2*b)^2 or 2(b^2)

if it was the first one when b=3 it will equal 36 which will give an even value of a becuase the square roots cancel

you might get better answers if you post this in the maths forum

Reply 2

DFranklin

18

JohnP

Apparently the Pythagoreans claim to have proved that the square root of two is irrational by examining the equation a^2 = 2*b^2. They observed that the square of any even number is even, and that of any odd number is odd, which I don’t have any difficulty with. However the proof requires reverse symmetry, as it claims that because a^2 is even then so must be a. How come?Well, a must be either even or odd, right?

But if it's odd, then by what you've already agreed with, its square must be odd, which contradicts the fact that a^2 is even.

So a can't be odd, so a must be even.

If you set b=3 then a^2 = 18. a^2 is certainly even, but the square root of 18 seems anything but either rational or even!

Which is why setting b = 3 doesn't make much sense.

What the proof is saying is that there are no integer values for a, b such that that equation can be true. The argument being "well, suppose a, b were integers so that the equation is true. Then we end up with a impossible conclusion".

So setting b=3 and saying "there's no integer value of a that makes sense" isn't an argument against the proof. Because that's what the proof is saying as well: that whatever value you choose for b, there's no integer value for a that makes sense.

Reply 3

Tallon

16

I think the whole point is that a fraction is wirrten with integers and so a^2 = 18 would never happen.
so if 2b^2 = a^2 then we know a is even.
if a is even it can at least be written as 2k instead.
2b^2 = (2k)^2
2b^2 = 4k^2
b^2 = 2k^2
therefore b^2 is even and therefore b is even too.

Reply 4

BJack

19

To continue what Tallon said, we start by saying that

\sqrt{2} = \frac{a}{b}

, where a and b are coprime integers.

Hence the conclusion that both a and b are even contradicts this statement, so the assertion that

\sqrt{2}

is rational must be false.

Reply 5

JohnP

OP

Are you saying that fractions must be even or odd? I had assumed only integers could be so.

Reply 6

around

13

What he's saying is that the two numbers involved in the fraction a/b are either even or odd.

Reply 7

refref

JohnP

Are you saying that fractions must be even or odd? I had assumed only integers could be so.

Well...

We first assume that root 2 is equal to a fraction, a / b where a and b have no common factor (a rational number is a number that is equal to a fraction a / b). That means we assume the fraction has already cancelled to be in its lowest terms. Take the fraction 3/6, this fraction isnt in its lowest terms, since they both have the factor 3 so we can cancel it to 1/2. Take a fraction with an even a and b, like 2/4. Since even numbers are all factors of 2, we can cancel the fraction to 1/2.

We assume that root 2 = a / b where a/b is a fraction like 1/2, 3/4, 7/8, 3/17, where the fraction is in its lowest terms.

In the proof we find that a is even, and then b is even. If a and b are both even, then they both have the factor 2. But we assumed that the fraction is in its lowest terms, but a fraction with an even a and b can be cancelled down, so our original assumption (that root 2 = a / b) must have been wrong.

Reply 8

nuodai

17

You begin by saying that

\sqrt{2} = \frac{a}{b}

, where

\frac{a}{b}

is a fraction in its simplest form. In other words, the lowest common divisor of

a

and

b

is 1. If, then, you show that it is a necessary condition for both

a

and

b

to be even, then we reach a contradiction, because in such a case the lowest common divisor is 2, not 1.

So, you've shown that

a^2 = 2b^2

, and since

a^2

is even,

a

must be even, so we can write

a

as

2k

.

Spoiler

Reply 9

JohnP

OP

Ok, folks, I think I am there now! My original reading that the proof requires reverse symmetry of the statement that if ‘a’ is even then a^2 is also even, i.e. that if a^2 is even then ‘a’ must also be even was a red herring and not directly required.

To summarise:

1). If the square root of 2 is rational and reduced to its lowest common denominator then ‘a’ and ‘b’ cannot both be even.

2). If ‘a’ is even then b^2 must be even (substitute a = 2k). But if b^2 is even then ‘b’ cannot be odd. So ‘a’ cannot be even, or if 'b' is neither then the solution cannot be rational.

3) If alternatively ‘a’ is odd then a^2 is odd. But we know a^2 is even as it = 2*b^2.

4) Thus ‘a’ cannot be either even or odd, and therefore the square root of two cannot be rational as first assumed.

Hope you all agree with this! Thanks for the discussion.

Reply 10

refref

JohnP

Ok, folks, I think I am there now! My original reading that the proof requires reverse symmetry of the statement that if ‘a’ is even then a^2 is also even, i.e. that if a^2 is even then ‘a’ must also be even was a red herring and not directly required.

To summarise:

1). If the square root of 2 is rational and reduced to its lowest common denominator then ‘a’ and ‘b’ cannot both be even.

2). If ‘a’ is even then b^2 must be even (substitute a = 2k). But if b^2 is even then ‘b’ cannot be odd. So ‘a’ cannot be even.

3) If alternatively ‘a’ is odd then a^2 is odd. But we know a^2 is even as it = 2*b^2.

4) Thus ‘a’ cannot be either even or odd, and therefore the square root of two cannot be rational as first assumed.

Hope you all agree with this! Thanks for the discussion.

Nearly there, just the order is a bit weird.

if ‘a’ is even then a^2 is also even, i.e. that if a^2 is even then ‘a’ must also be even was a red herring and not directly required.

This is an important part.

If (a^2) is even (we are only talking about integers here) then a must also be even. This is because a^2 = a*a, and for a^2 to be even, then one of its factors must be even, namely just a, so a is even.

\sqrt{2} = \frac{a}{b}

2 = \frac{a^2}{b^2}

2b^2 = a^2

The last line is important. Forget the right side as being a^2 and just treat it as any number, call it Z (you dont have to do this, its just to help you understand). Now 2b^2 = Z, 2 times a number (the number in this case is b^2) equals an even number, so Z (or a^2) must be even, so 'a' must be even, so a = 2k so a^2 = (2k)^2 = 4k so b^2 = 2k so by the same arguement, b is also even, so the original fraction couldnt have been in its lowest terms.

You dont really need to ever talk about a being odd, as we have shown that it is even.

Proof that sq root of 2 is irrational

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