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

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• Nov 30th 2013, 07:17 PM
Melody2
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.
-------------------------------------------------
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).
• Nov 30th 2013, 07:45 PM
romsek
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.
-------------------------------------------------
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?
• Nov 30th 2013, 07:57 PM
Melody2
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?
• Nov 30th 2013, 08:30 PM
romsek
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.
• Nov 30th 2013, 08:48 PM
Melody2
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.
• Nov 30th 2013, 08:51 PM
romsek
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.
• Nov 30th 2013, 09:03 PM
Melody2
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
• Nov 30th 2013, 09:18 PM
romsek
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.
• Nov 30th 2013, 09:24 PM
Melody2
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
• Nov 30th 2013, 09:28 PM
romsek
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
• Nov 30th 2013, 10:14 PM
Melody2
Re: Prove (square root 2) is irrational - I have a problem with this proof.

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
--------------------------------------------------
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.
----------------------------------------------------

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.
---------------------------------------------------------
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.
• Dec 1st 2013, 12:45 AM
romsek
Re: Prove (square root 2) is irrational - I have a problem with this proof.
Glad you got it squared away.

Romsek (with a k!)
• Dec 1st 2013, 04:57 AM
Plato
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.

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 $\displaystyle K$.
• Dec 1st 2013, 06:44 PM
Melody2
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.
• Dec 1st 2013, 08:41 PM
romsek
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!

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