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

In summary, the conversation discusses a proof that √3 is irrational, with the person sharing their own proof and questioning its validity. The main issue is that the proof assumes p/q is an odd integer, but it is not known if p/q is even an integer to begin with. The conversation ends with the person realizing their mistake and going back to revise their proof.
  • #1
mindarson
64
0
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.
 
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  • #2
Your logic is invalid. How do you know p/q is, let alone odd, an integer?
 
  • #3
p/q is in general a rational, but you define it's parity. How does that work?
 
  • #4
"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
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
mindarson said:
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
micromass said:
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.
 

1. Why is it important to prove that Sqrt[3] is irrational?

Proving that Sqrt[3] is irrational has significance in mathematics as it helps establish the existence of irrational numbers and their properties. It also serves as a foundation for more complex proofs and theorems.

2. How can we prove that Sqrt[3] is irrational?

One way to prove that Sqrt[3] is irrational is by contradiction. Assume that Sqrt[3] is rational, and then show that this leads to a contradiction. This will prove that the initial assumption was false, and hence Sqrt[3] must be irrational.

3. Can you explain the logic behind the proof of Sqrt[3] being irrational?

The proof of Sqrt[3] being irrational follows the idea that if a number can be written as a ratio of two integers, then it is rational. However, by assuming that Sqrt[3] is rational and using basic algebraic manipulations, we arrive at a contradiction, thus proving the initial assumption false.

4. Is there a simpler way to understand the proof of Sqrt[3] being irrational?

The proof of Sqrt[3] being irrational can also be understood through the concept of prime factorization. By showing that the square root of 3 cannot be written as a ratio of two integers with no common factors, we can conclude that it is irrational.

5. Can the proof of Sqrt[3] being irrational be applied to other square roots?

Yes, the same proof can be applied to any square root of a non-perfect square number. This is because the fundamental logic behind the proof remains the same, only the numbers involved may change. However, this does not apply to all irrational numbers as some may have different proofs of their irrationality.

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