Prove/disprove that $\sqrt 3$ is irrational

Question

zugzwang · Answer

Things like this are just begging to be proven by contradiction...

anonymous · Answer

it's irrational by definition

zugzwang · Answer

Just go for assuming that it's rational, and then look for a contradiction...

lgbasallote · Answer

taking the easy way out huh @nphuongsun93

anonymous · Answer

:D.. *poke google* < http://en.wikipedia.org/wiki/Square_root_of_3#Proof_of_irrationality >

lgbasallote · Answer

$$\sqrt 3 = \frac ab$$
where a/b is in simplest terms
$$3 = \frac{a^2}{b^2}$$
$$3b^2 = a^2$$
then?

lgbasallote · Answer

@nphuongsun93 don't spoil the steps

zugzwang · Answer

Well, if a and b are relatively prime, what can you say about a^2 and b^2?

anonymous · Answer

depends on what you can use
easiest way is to assume 
$$3a^2=b^2$$ then by the fundamental theorem of arithmetic, factor $a^2$ and\  $b^2$ uniquely into products of primes
observe the right hand side has an even number of factors of 3 whereas the left hand side has an odd number of them

lgbasallote · Answer

i can say that a and b cannot both be even...

zugzwang · Answer

Their squares must also be relatively prime, right?

lgbasallote · Answer

by relatively prime you mean...? i'm not really a math lover like you (as you can tell from my profile picture)

zugzwang · Answer

Relatively prime, their greatest common factor is 1.
It's a consequence of a/b being in simplest terms..

lgbasallote · Answer

so a and b cannot both be even....

lgbasallote · Answer

same meaning......

zugzwang · Answer

That follows from them being relatively prime.

lgbasallote · Answer

as long as they're same...the end justifies the means

zugzwang · Answer

Well, there you have it, then.

lgbasallote · Answer

hmm... i think i see the contradiction now

lgbasallote · Answer

a and b cannot both be even

so if b is odd

3b^2 is odd

but 3b^2 = a^2 and a^2 is even

odd = even <--contradiction

if a is odd

3b^2 is even

3b^2 = a^2 and a^2 is odd

even = odd same thing

zugzwang · Answer

uhh...
a and b cannot both be even OK
if b is odd, then 3b^2 is odd OK
3b^2 = a^2, a^2 is even... No it isn't... you don't know what it is yet.

lgbasallote · Answer

i just said a and b cannot both be even...so if b is odd, then a is even. if a is even then a^2 is even....

zugzwang · Answer

If b is odd, then a is even?  I think you just proved that if a is odd, then b is odd,

lgbasallote · Answer

uhh what??

lgbasallote · Answer

i just said that a and b can't both be even...when did i say they're both odd?

zugzwang · Answer

You didn't. You didn't say that if b is odd, then a is even.

lgbasallote · Answer

i said... a and b can't both be even...in other words...either a is even and b is odd...or a is odd and b is even

lgbasallote · Answer

that was the definition of rational number

a/b where a and b are integers and a and b are not both even (because if it's even then it's factorable and not in lowest terms)

zugzwang · Answer

Look carefully
a and b are not both even, in symbols, this is
~(a is even AND b is even)
(a is odd OR b is odd)
it is not the same as 
(a is even OR b is even)

zugzwang · Answer

In other words, it can be the case that both are odd.

lgbasallote · Answer

yes....i wasn't there yet....i was just proving case 1 and 2

lgbasallote · Answer

you instantly called my case 1 wrong already

zugzwang · Answer

In any case, if a and b are relatively prime, then so are a^2 and b^2
but a^2/b^2 = 3, then a^2 = 3b^2
Then
a^2/b^2 = 3b^2/b^2... means a^2 and b^2 are not relatively prime, which is a contradiction...

lgbasallote · Answer

i just said don't spoil the steps....

anyway..you've done it....so how a^2/b^2 = 3b^2/b^2 are not relatively prime?

zugzwang · Answer

Well, b^2 is a common factor

lgbasallote · Answer

?

zugzwang · Answer

Scratch that proof, I think it's inaccurate.

lgbasallote · Answer

maybe $\sqrt 3$ isn't irrational

zugzwang · Answer

lol it is
We were pretty much on the right track...
Ok, you've ruled out the possibility that they're both even.
What about the other cases?

zugzwang · Answer

By both, I mean both a and b.

zugzwang · Answer

Suppose a is odd and b is even.
then a^2 is odd and b^2 is even
3b^2 is even, but it is equal to a^2, an odd number, so this cannot be.
Suppose a is even and b is odd
Then a^2 is even and b^2 is odd
3b^2 is odd, but it is equal to a^2, an even number, so this cannot be.

Hence we are left with the case that both a and b must be odd.

zugzwang · Answer

I realise you did that earlier... oh well...

zugzwang · Answer

Sorry for causing delay ... >.>

lgbasallote · Answer

hmm what about this

a and b are both odd so...
$$3(2k+1)^2 = (2x+1)^2$$
$$3(4k^2 + 4k + 1) = 4x^2 + 4x + 1$$
$$12k^2 + 12k + 3 = 4x^2 + 4x + 1$$
$$12k^2 + 12k - 4x^2 + 4x = -2$$
$$4(3k^2 + 3k - x^2 + x) = -2$$
$$3k^2 + 3k - x^2 + x = -\frac 12$$
contradiction

zugzwang · Answer

Something like that...
Except in the third line, it should be -4x.
Just make sure to indicate exactly why it must be a contradiction, and you're done.

zugzwang · Answer

Sorry, fourth line, not third.

lgbasallote · Answer

still contradicts though

lgbasallote · Answer

since you ahve the most repy in this thread, i'll medal you

zugzwang · Answer

lol, thanks :)