Two "valid" proofs that sqrt(2) is irrational I think have flaws in it.

[Steven G]

Claim: sqrt(2) is irrational
Proof: Suppose otherwise. That is assume sqrt(2) = a/b where gcd(a,b)=1
Then a^2=2b^2=(2b)b =>b|a^2
If b>1, then the Fundamental Theorem of Arithmetic guarantees that there exists a prime p such that p|b.
Since p|b and b|a^2 => p|a^2. That is p|a*a. Well p has to divide (at least) one of those two values (a or a). So p|a (and p|b).
This means that gcd(a,b)>=p. This is a contradiction.
Now if b=1, then a^2=2. There is no integer that when we multiply it by itself gives us the number 2.
As a result of the above, sqrt(2) is irrational.
How can we assume that there is no integer that when we multiply it by itself gives us the number 2?

I have trouble with the more conventional proof as well.
Proof: Suppose sqrt(2) = a/b where a/b is in reduced form.
Then a^2=2b^2 => a ^2 is even => a is even.
Now let a=2r=>a^2 = 4r^2.
Then 4r^2 =2b^2 or 2r^2 = b^2 =>b^2 is even=> b is even.
If and b are both even, then a/b can't be in reduced form.
How do we know that a/b can be written in reduced form?

 

[Dr.Peterson]

How can we assume know that there is no integer that when we multiply it by itself gives us the number 2?

Because 1^2 = 1 and 2^2 = 4. There is no integer between them.

How do we know that a/b can be written in reduced form?

That would be an earlier theorem, related to the gcd.

I presume you recognize that the first proof starts equivalently, in saying "assume sqrt(2) = a/b where gcd(a,b)=1" which is the same as "Suppose sqrt(2) = a/b where a/b is in reduced form".

 

[Steven G]

Because 1^2 = 1 and 2^2 = 4. There is no integer between them.


That would be an earlier theorem, related to the gcd.

I presume you recognize that the first proof starts equivalently, in saying "assume sqrt(2) = a/b where gcd(a,b)=1" which is the same as "Suppose sqrt(2) = a/b where a/b is in reduced form".

Because 1^2 = 1 and 2^2 = 4. There is no integer between them. I was hoping that you would be one of the helpers that answered this post. It really is enough just to say 1^2=1 and 2^2=4? It doesn't have to be more rigorous. I have to accept what you say as you're the expert.

How do we know that a/b can be written in reduced form?

That would be an earlier theorem, related to the gcd.
Can you please show me this proof? I thought that it might come from the well ordering principal.

I presume you recognize that the first proof starts equivalently, in saying "assume sqrt(2) = a/b where gcd(a,b)=1" which is the same as "Suppose sqrt(2) = a/b where a/b is in reduced form".
I agreed with the about statement somewhat and here is why. We can show that the gcd(a,b)=1 but I am not convinced yet that we can write a/b in reduced form. But if we could, then the statement are equivalent.

 

[Steven G]

Oh right, now I see the theorem that I needed. I recently proved that if gcd(a,b)=d, then gcd(a/d, b/d) = 1. Got it. Thanks!

 

[lookagain]

I thought that it might come from the well ordering principal.

It is the "well-ordering principle."


Neither the principal nor the vice-principal do the well-ordering.\(\displaystyle \ \ \) : )

 

[Dr.Peterson]

It really is enough just to say 1^2=1 and 2^2=4? It doesn't have to be more rigorous. I have to accept what you say as you're the expert.

Did I say I was writing a complete proof? No, if you are writing an actual proof, you will want to say more, such as that squaring is an increasing function, and probably say it more formally (unless perhaps it is a theorem in your book).

But in my mind, this is the key idea, and if someone doesn't say it explicitly, it would be because it is considered to be well-known or "obvious".

And, no, I'm not the expert. (I never actually took, much less taught, a number theory course.) And if I were, who ever said that math depended on authority?

We can show that the gcd(a,b)=1 but I am not convinced yet that we can write a/b in reduced form. But if we could, then the statement are equivalent.

No one is showing the gcd is 1; that is being assumed!

But to me, the definition of "reduced form" is that gcd(a,b) is 1. That's what I mean by equivalent. Both proofs start by assuming that any fraction can be written in this form, which is valid because of a prior theorem.

 

[pka]

Here is a useful theorem(one & done): If [imath]p\in\mathbb{Z}^+[/imath] that is not a square then [imath]\sqrt{p}[/imath] is irrational.
To prove the theorem, suppose that [imath]\sqrt{p}[/imath] is rational. Then there are two positive integers [imath]a~\&~b[/imath] such that [imath]\sqrt{p}=\dfrac{a}{b}[/imath] or [imath]b\sqrt{p}=a[/imath].
Let [imath]\{n\in\mathbb{Z}^+: n\sqrt{p}\in\mathbb{Z}^+\}[/imath] It is clear that [imath]b\in T[/imath] so [imath]T\ne\emptyset[/imath].
Every set of positive integers contains its [imath]GLB[/imath] Then suppose that [imath]t= GLB(T)[/imath]
Thus [imath]t\in T\text{ or }t\cdot\sqrt{p}\in\mathbb{Z}^+[/imath]. Let [imath]H = \left\lfloor {\sqrt p } \right\rfloor [/imath], the greatest integer not greater than [imath]\sqrt{p}[/imath].
We know from the properties of the floor function [imath]0<\sqrt{p}-H<1[/imath]. That implies that [imath]0<t\sqrt{p}-tH<t[/imath]
BUT that means [imath]\bf{(t\sqrt{p}-tH)\sqrt{p}}[/imath] is also an integer. that is a contradiction, to what?
[imath][/imath][imath][/imath][imath][/imath]