But we assumed p and q are mutually prime, a contradiction.

[rohanprabhu]

I read this on cut-the-knot.org:

The premise $p^2 = 2q^2$ tells us that p is even. Assuming p and q mutually prime, q is bound to be odd. However, the square of an even number is divisible by 4, which leads us to conclude that q is even. A contradiction.

How is this a contradiction? If $p^2 = 2q^2$ just says that $p^2$ is even, it doesn't say that 'p' is even. It is a very obvious mistake. So.. is this a stupidity on their part or something that i may have missed [or a stupidity] on my part..

I generally wouldn't have posted this here.. but since cut-the-not.org has quality articles posted.. i would like to rethink this problem before concluding that they made a mistake.

thanks

 

[CaptainQuasar]

Is there an assumption that p and q are integers? I think that might make it work… but I agree with your objection.

⚛​

 

[slider142]

The problem assumes that p and q are integers, since it assumes that p/q is a rational number.

 

[CaptainQuasar]

The problem assumes that p and q are integers, since it assumes that p/q is a rational number.

Not all rational numbers are integers.

$$\frac{1}{2}$$

is a rational number but not an integer.

⚛​

 

[Gib Z]

Slider142 never said p/q was an integer if p and q were integers, but did say p/q was rational if p and q were integers. 1 and 2 are nice integers =]

We have that p is an integer, and that p^2 is divisible by 2. In fact, p^2, if divisible by 2, must also be divisble by 4. This is because if we express p^2 as a product of its factors, $$p^2 = 2^2 * k^2 * p^2 ...$$ where k and p are other integers/factors. All the powers must be even, in particular for 2. This is because if the power were odd, then in the factorization for p, there would be a non-integer power - ie not allowed.

 

[CaptainQuasar]

Slider142 never said p/q was an integer if p and q were integers, but did say p/q was rational if p and q were integers. 1 and 2 are nice integers =]

Oh, you're right. Duh. Sorry, slider.

⚛​

 

[HallsofIvy]

If p is odd, then p= 2n+1 for some integer n. Then p²= (2n+1)²= 4n²+ 4n+ 1= 2(2n²+ 2n)+ 1, "2 times an integer plus 1", and so odd. That is: If p is an odd integer, then so is p²". It follows that if p² is even then p must be even.

I would suggest that you not throw around words like "obviously wrong" and "stupidity" so easily. They tend to come back at you.

 

[ramsey2879]

I read this on cut-the-knot.org:



How is this a contradiction? If $p^2 = 2q^2$ just says that $p^2$ is even, it doesn't say that 'p' is even. It is a very obvious mistake. So.. is this a stupidity on their part or something that i may have missed [or a stupidity] on my part..

I generally wouldn't have posted this here.. but since cut-the-not.org has quality articles posted.. i would like to rethink this problem before concluding that they made a mistake.

thanks

They were talking oranges but you are talking bananas. The contradiction is that p and q were originally considered coprime! Since if the square root of 2 = p/q where p and q are integers, we can always reduce p and q to coprime integers by dividing by the common factor.

 

[al-mahed]

it's impossible two number be equal and one be even and the other be odd, and is also impossible a square of an odd/even number turn to be even/odd

 

[ramsey2879]

it's impossible two number be equal and one be even and the other be odd, and is also impossible a square of an odd/even number turn to be even/odd

But 2*p is even, even if p is odd, so I don't know what you are trying to say.

 

[al-mahed]

But 2*p is even, even if p is odd, so I don't know what you are trying to say.

I am saying that if x = 2y then x is even, very simple.

 

[HallsofIvy]

it's impossible two number be equal and one be even and the other be odd, and is also impossible a square of an odd/even number turn to be even/odd

But 2*p is even, even if p is odd, so I don't know what you are trying to say.

Yes 2p is always even but that has nothing to do with what he said. He, and your original post, was talking about squares of numbers. As I showed above, the square of any even number is even, the square of any odd number is odd.

If n² is even, then n itself must be even. If it were odd, then n² would have to be odd, not even.

 

[ramsey2879]

I am saying that if x = 2y then x is even, very simple.

That is fine but Can you follow HallsofIvy's Logic and show that If x is square then y can not be odd? Thus the contradictiopn since y was assumed to be odd?

 

[al-mahed]

Hi Ramsey, I am not sure what you want to know about my post...

given x = 2y, if y is odd then x cannot be a square since x is even.

proof: if x is a square it can be written as 4a^2, for an any a, and 4a^2 = 2y ==> 2a^2 = y ==> y is even, and this is a contradiction because we assumed that y is odd


I am not sure if this is what you want, so I'll reproduce a proof of irrationality of 2

supose $$\sqrt{2} = p/q$$ in the lowest terms ==> gcd(p,q) = 1

$$2 = \frac{p^2}{q^2}\ ==> \ 2q^2 = p^2\ ==>$$ p is even (this is the conclusion about the elementary principle: two numbers cannot be equal and one be even and the other be odd, and I am not talking about q, I am talking about 2q^2, this number is even)

as p is even, p = 2k ==> p^2 = 4k^2 ==> $$q^2 = 2k^2$$ which means that q is even also, so p/q is not in the lowest terms, and this is the contradiction, and then $$\sqrt{2}$$ cannot be expressed as a rational ==> $$\sqrt{2}$$ is irrational

 

[ramsey2879]

Hi Ramsey, I am not sure what you want to know about my post...

given x = 2y, if y is odd then x cannot be a square since x is even.

proof: if x is a square it can be written as 4a^2, for an any a, and 4a^2 = 2y ==> 2a^2 = y ==> y is even, and this is a contradiction because we assumed that y is odd


I am not sure if this is what you want, so I'll reproduce a proof of irrationality of 2

supose $$\sqrt{2} = p/q$$ in the lowest terms ==> gcd(p,q) = 1

$$2 = \frac{p^2}{q^2}\ ==> \ 2q^2 = p^2\ ==>$$ p is even (this is the conclusion about the elementary principle: two numbers cannot be equal and one be even and the other be odd, and I am not talking about q, I am talking about 2q^2, this number is even)

as p is even, p = 2k ==> p^2 = 4k^2 ==> $$q^2 = 2k^2$$ which means that q is even also, so p/q is not in the lowest terms, and this is the contradiction, and then $$\sqrt{2}$$ cannot be expressed as a rational ==> $$\sqrt{2}$$ is irrational

The key was the assumption that gcd(p,q) = 1 which is what Slider142 and I posted earlier.

 

[al-mahed]

oh yes, ramsey, this is the key.

sometimes I just read the first post and put some ideas, I didin't see what you guys have posted before

I hope who have opened the thread do understand the arguments

 

[alphachapmtl]

$p^2 = 2q^2$
so $p^2$ is even
so $p$ is even
so $p$=0 mod 2
so $p^2$=0 mod 4
so $p^2=2q^2$=0 mod 4
so $q^2$=0 mod 2
so $q$=0 mod 2
so q is even
so 2|gcd(p,q)
so gcd(p,q)>1