Logical dilemma involving the proof of the irrationality of sqrt(2).

I understand the proof, through and through. However, I have a logical dilemma involving the proofs conclusion.

The proof doesn't seem valid to me. I looked around the web and saw very similar

descriptions for this proof and they too all seem invalid to me. All the proofs start with the idea that a rational number can be written as the ratio of two integers, say a/b, and that for any ratio there exists exactly one fully reduced fraction (where no integer greater than 1 exists that can be evenly divided into both the numerator and denominator.)

What I see is there may exist a fraction that is not fully reduced. Even if I assumed a non-fully reduced fraction did exist, this does not imply to me there does not exist a non-fully reduced fraction.

In other words, just because you found a solution that proves that the solution itself is not in reduced/lowest terms; it doesn't mean that the solution couldn't possibly be not in reduced lowest terms? I also realized that we assumed that it was in reduced lowest terms.

However, that's like assuming that X is a natural number and 2x + 5 = 6. So you proved that it isn't a natural number, contradicting your assumption - so what?

Re: Logical dilemma involving the proof of the irrationality of sqrt(2).

Quote:

Originally Posted by

**GDM** I understand the proof, through and through. However, I have a logical dilemma involving the proofs conclusion.

The proof doesn't seem valid to me. I looked around the web and saw very similar

descriptions for this proof and they too all seem invalid to me. All the proofs start with the idea that a rational number can be written as the ratio of two integers, say a/b, and that for any ratio there exists exactly one fully reduced fraction (where no integer greater than 1 exists that can be evenly divided into both the numerator and denominator.)

What I see is there may exist a fraction that is not fully reduced. Even if I assumed a non-fully reduced fraction did exist, this does not imply to me there does not exist a non-fully reduced fraction.

In other words, just because you found a solution that proves that the solution itself is not in reduced/lowest terms; it doesn't mean that the solution couldn't possibly be not in reduced lowest terms? I also realized that we assumed that it was in reduced lowest terms.

However, that's like assuming that X is a natural number and 2x + 5 = 6. So you proved that it isn't a natural number, contradicting your assumption - so what?

**Every non-fully reduced fraction can be reduce to fully reduced fraction!**

Re: Logical dilemma involving the proof of the irrationality of sqrt(2).

Quote:

Originally Posted by

**GDM** I understand the proof, through and through. However, I have a logical dilemma involving the proofs conclusion.

The proof doesn't seem valid to me. I looked around the web and saw very similar

descriptions for this proof and they too all seem invalid to me. All the proofs start with the idea that a rational number can be written as the ratio of two integers, say a/b, and that for any ratio there exists exactly one fully reduced fraction (where no integer greater than 1 exists that can be evenly divided into both the numerator and denominator.)

What I see is there may exist a fraction that is not fully reduced. Even if I assumed a non-fully reduced fraction did exist, this does not imply to me there does not exist a non-fully reduced fraction.

In other words, just because you found a solution that proves that the solution itself is not in reduced/lowest terms; it doesn't mean that the solution couldn't possibly be not in reduced lowest terms? I also realized that we assumed that it was in reduced lowest terms.

However, that's like assuming that X is a natural number and 2x + 5 = 6. So you proved that it isn't a natural number, contradicting your assumption - so what?

As was said, every fraction can be written in a simplest form if it isn't already. So it is fine to define $\displaystyle \displaystyle \sqrt{2} = \frac{a}{b}$ to say that $\displaystyle \displaystyle \frac{a}{b}$ is already in its simplest form, since you have made that assumption by saying the number is rational. When you are able to "simplify" it, you arrive at your contradiction.

Re: Logical dilemma involving the proof of the irrationality of sqrt(2).

I want to take the opportunity to mention Conway's lecture presenting a superb proof of irrationality of $\displaystyle \sqrt{2}$, Morley's theorem and something from knot theory.

Here: http://www.cs.toronto.edu/~mackay/conway.pdf

Re: Logical dilemma involving the proof of the irrationality of sqrt(2).

Thanks for the information, and the links.

Re: Logical dilemma involving the proof of the irrationality of sqrt(2).

the idea is, that if we have a fraction a/b, and d is a common divisor of a and b, then

a = a'd

b = b'd.

we then get a new fraction a'/b' which equals the old one, since: ab' = (a'd)b' = a'(db') = a'(b'd) = a'b

(this is how we determine equality of fractions: we "cross-multiply". for example, 3/4 = 6/8 because (3)(8) = 24 = (4)(6).).

if d is the greatest common divisor, then a'/b' is "fully reduced" (convince yourself that this reduced fraction is unique).

Re: Logical dilemma involving the proof of the irrationality of sqrt(2).

I'll take a stab at explaining the logical dilemma you describe. Since squares of odd numbers are odd and squares of even numbers are divisible by 4, we can make the following argument: if $\displaystyle \frac a b=\sqrt2$, then $\displaystyle a^2=2b^2$. Clearly $\displaystyle a^2$ is even, and therefore divisible by 4. Dividing both sides by 2, we get that $\displaystyle b^2$ is even. So both a and b are even.

So far, all we've proved is that any fraction equal to the square root of 2 must have even numerator and denominator, which I think is what you were saying in your original post. But dividing numerator and denominator by 2 does not change the number (both fractions represent the *same* rational number). And we can't keep dividing numerator and denominator by 2 forever, because there has to be some k such that 2^k is larger than the denominator. So after at most k steps of dividing the top and bottom by 2, we get a fraction with odd numerator or denominator, but which is equal to the square root of 2, and that's our contradiction.

The appeal to "lowest terms" at the beginning of many proofs is clumsy, IMO, because first it just assumes that the integers have unique factorization, and second, that's a stronger property than is required to prove the result.