Proof that Sqrt[3] is irrational - Is my logic valid?

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I am self-studying elementary analysis and am learning how to prove things. I have come up with a proof that √3 is irrational, and I believe it is valid, but I am unsure of my logic, as I have not seen it proved in just this way, and I don't have a prof to ask! So if anyone could just take a look at my argument and tell me if it is valid, I would appreciate it very much.

Here goes.

Theorem. √3 is irrational, i.e., there is no pair of integers p,q (q ≠ 0) satisfying (p/q)2 = 3.

Proof (by contradiction). Assume, for contradiction, that there are two integers p,q (q≠ 0) such that

(p/q)2 = 3

Then p/q is odd (it is easily proved by contradiction that if n2 is odd, then n must be odd for all integers n), i.e., there is an integer n such that

p/q = 2n + 1

and therefore

p = q(2n + 1).

This can be substituted into the first equation, giving

(2n + 1)2 = 3

which expands to

2n(2n + 1) + 2n + 1 = 3 → 4n2 + 4n = 2 → 2(n2 + n) = 1

Since n2 + n is an integer (closure under addition and multiplication), this last statement implies that 1 is an even integer, which it is not. Therefore there is no pair of integers p,q (q ≠ 0) satisfying (p/q)2 = 3, i.e., √3 is irrational, which is what was required to be shown.

That's the proof I came up with. My specific question is this: Most proofs of this theorem are similar to the proof of the irrationality of √2, i.e. they start by assuming p and q have no common factor and end by deriving a contradiction of this assumption. My proof does not derive a contradiction of a hypothesis but instead derives a contradiction of a known fact, i.e. that 1 is an odd integer. Is this a 'legal' move, or not?

Thanks for taking time to help.
 

Answers and Replies

  • #2
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Your logic is invalid. How do you know p/q is, let alone odd, an integer?
 
  • #3
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p/q is in general a rational, but you define it's parity. How does that work?
 
  • #4
HallsofIvy
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"Even" and "0dd" work nicely when proving that [itex]\sqrt{2}[/itex] is not rational because "even" and "odd" depend upon 2.

What you need is "if [itex]p^2[/itex] is a multiple of 3, then p is a multiple of 3". Can you prove that?
 
  • #5
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I'm sorry, but I'm not convinced that my logic is invalid. I guess I am misunderstanding your objections. (I actually don't even know what 'parity' is!)

The main problem seems to be that I am using the fact that p/q must be an odd integer. I know that p and q are integers because I stipulated they were at the outset. I know that p/q must be odd because (p/q)2 must be odd, if it is to equal 3, which I am assuming it does, since 3 is itself an odd integer. And if n2 is odd, then n is odd perforce.

From this oddness of (p/q) I derive a patent absurdity by simple algebra. And since my assumption led directly to an absurdity, my assumption must have been wrong. I just don't see where this reasoning fails.

Unless... perhaps I am making more than one vulnerable assumption, and my proof leaves me no way to decide which one to jettison. Is this what is meant by "How do I know p and q are integers?" perhaps?
 
  • #6
micromass
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The main problem seems to be that I am using the fact that p/q must be an odd integer.

How do you know p/q is an integer?? All you know is that p and q are integers, this only implies that p/q is rational.

For example, we could have p/q=1/2. Is this an odd integer?? It doesn't even make sense.
 
  • #7
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For example, we could have p/q=1/2. Is this an odd integer?? It doesn't even make sense.

Wow. Facepalm of my life. Sorry I wasted everyone's time with such a dumb oversight. But, as the great RP McMurphy would say:

"But I tried, didn't I? Goddammit, at least I did that."

Back to the drawing-board.
 

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