Prove (square root 2) is irrational - I have a problem with this proof.

I believe that this is a standard proof but I have a problem with it.

4d383176cd06283c92755eafb4b9586c.gif

The end statement appears to me to be

root2 = a/b where a and b are both even therefore a and b are not coprime which contradicts the initial condition that a and b must be coprime therefore root2 is irrational.

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I ran through a different proof to attempt to prove that root4 is irrational

(this is not the full proof, I have truncated it a little)

suppose root4 were irrational

then root4 = a/b

4b^2=a^2

therefore a is even

Let a=2k

4b^2=4k^2

b^2=k^2

k=+-b

therefore

a=2k and b=+/-k

Therefore a and b are not coprime

Therefore root4 is irrational.

Obviously I realize that 2k/k can be reduced to 2/1 which then shows root4 to be rational

BUT

when the first proof was looking at root2 and it got down to (an even number/an even number) so why is it impossible for that fraction also be reduced to some p/q where p and q are coprime. To me, the proof just doesn't seem to be finished and does not prove anything (not to me).

Re: Prove (square root 2) is irrational - I have a problem with this proof.

Quote:

Originally Posted by

**Melody2** I believe that this is a standard proof but I have a problem with it.

4d383176cd06283c92755eafb4b9586c.gif
The end statement appears to me to be

root2 = a/b where a and b are both even therefore a and b are not coprime which contradicts the initial condition that a and b must be coprime therefore root2 is irrational.

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I ran through a different proof to attempt to prove that root4 is irrational

(this is not the full proof, I have truncated it a little)

suppose root4 were irrational

then root4 = a/b

4b^2=a^2

therefore a is even

Let a=2k

4b^2=4k^2

b^2=k^2

k=+-b

therefore

a=2k and b=+/-k

Therefore a and b are not coprime

what if b=1, as it does here?

Re: Prove (square root 2) is irrational - I have a problem with this proof.

Thanks but I don't get what you are trying to tell me Romsek.

b=+/-k and

k can equal any whole number it doesn't have to be 1. (or does it?)

Does the answer have implied restrictions on k?

Re: Prove (square root 2) is irrational - I have a problem with this proof.

Quote:

Originally Posted by

**Melody2** Thanks but I don't get what you are trying to tell me Romsek.

b=+/-k and

k can equal any whole number it doesn't have to be 1. (or does it?)

Does the answer have implied restrictions on k?

you used the fact that a=2k and b=+/-k to say that a and b are not coprime. But this isn't true if b=k=1.

Re: Prove (square root 2) is irrational - I have a problem with this proof.

Ok so you are telling me that they must be coprime for some particular value of k

whereas I was thinking that they must be coprime for all values of k.

I guess that helps. Thanks.

Re: Prove (square root 2) is irrational - I have a problem with this proof.

no, it just doesn't apply if k=1. Take 2 and 1. you can't reduce 2/1 any further. I don't know that you'd call 2 and 1 coprime but certainly the fraction 2/1 is in fully reduced form. Here k=1.

Re: Prove (square root 2) is irrational - I have a problem with this proof.

Quote:

Originally Posted by

**romsek** no, it just doesn't apply if k=1. Take 2 and 1. you can't reduce 2/1 any further. I don't know that you'd call 2 and 1 coprime but certainly the fraction 2/1 is in fully reduced form. Here k=1.

It was sqrt4 = a/b=2k/k if k is anything other than 1 then 2k and k are not coprime.

so this particular proof only works if k=1

Re: Prove (square root 2) is irrational - I have a problem with this proof.

Quote:

Originally Posted by

**Melody2** It was sqrt4 = a/b=2k/k if k is anything other than 1 then 2k and k are not coprime.

so this particular proof only works if k=1

the key to the proof of sqrt[2] being irrational though was that b=2k and was thus even. Which since a was found to be even earlier gives a contradiction that a/b was in fully reduced form.

you can't make that statement here.

Re: Prove (square root 2) is irrational - I have a problem with this proof.

The whole reason that that the factor of 2 was relevant was that it meant a and b were not coprime.

so although I hear frustration in your post I am not going to give way completely. Sorry

Re: Prove (square root 2) is irrational - I have a problem with this proof.

right but in your proof b = +/- k not +/- 2k, so you can't claim b is even here like the other proof did. No frustration here and I certainly don't view it as a fight :P

Re: Prove (square root 2) is irrational - I have a problem with this proof.

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We don't know exactly what k is __although it does have a specific value__. We do know that 'a' has a factor of 2 so therefore 'b' cannot be even.

etc

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If I make this change to the wording of the proof then I am happy and I think it all makes sense.

Thankyou for your help Romsec. I have enjoyed our communications this afternoon and I appreciate your help.

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I just a had a brainwave. Of course k has a specific value because 'a' has to have a specific value and k is half of a.

Now I have it. In the sqrt4 proof k has to be 1, there is not other possibliliy.

Thankyou so much for helping me reach this understanding.

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I know you kept telling me that k=1 but the real reason for this was not sinking into my brain.

Sometimes students (that's me) can be very frustrating. I didn't mean any offence by the comment.

Re: Prove (square root 2) is irrational - I have a problem with this proof.

Glad you got it squared away.

Romsek (with a k!)

Re: Prove (square root 2) is irrational - I have a problem with this proof.

Quote:

Originally Posted by

**Melody2**

I realize that you have solved this for P=2. But is is a general proof.

Suppose that $\displaystyle P$ is** a positive integer that is **__not a square__. We prove that $\displaystyle \sqrt P$ is irrational.

For a contradiction suppose that $\displaystyle \sqrt P$ is rational. Then the is a smallest positive integer $\displaystyle K$ for which it is true that $\displaystyle K\sqrt P$ is a positive integer.

One of the properties of the floor function is that $\displaystyle 0<\sqrt P-\left\lfloor {\sqrt P } \right\rfloor < 1 $

But that means $\displaystyle 0<K\sqrt P-K\left\lfloor {\sqrt P } \right\rfloor < K $ which means $\displaystyle K\sqrt P-K\left\lfloor {\sqrt P } \right\rfloor $ is a positive integer smaller that $\displaystyle K$

but $\displaystyle (K\sqrt P-K\left\lfloor {\sqrt P } \right\rfloor)\sqrt P $ is also a positive integer.

That contradicts the minimal nature of $\displaystyle K$.

Re: Prove (square root 2) is irrational - I have a problem with this proof.

Thanks Plato,

I at least followed your argument right up to the last line.

It then dissipated into the realm of mystery for me.

also,

how do you implant things like you implanted the 'floor function'.

I had seen these floor and ceiling symbols before but i had no idea what they were.

Thanks.

Re: Prove (square root 2) is irrational - I have a problem with this proof.

Quote:

Originally Posted by

**Melody2** Thanks Plato,

I at least followed your argument right up to the last line.

It then dissipated into the realm of mystery for me.

also,

how do you implant things like you implanted the 'floor function'.

I had seen these floor and ceiling symbols before but i had no idea what they were.

Thanks.

Embedded TeX can do all sorts of things!

$\displaystyle \lceil y\rceil \lfloor x\rfloor$