What is flawed in the reasoning about solving irrational numbers?

[glebovg]

Can anyone explain what is wrong with my reasoning?

Suppose x = \frac{p}{q} and let x = \sqrt 2 + \sqrt 3. Also, let a,b,c \in {\Bbb Z} and assume a < xc < b. If I show that xc must be an integer, and I know there does not exist c such that \sqrt 2 c, or \sqrt 3 c is an integer. Then, \left( {\sqrt 2 + \sqrt 3 } \right)c cannot be an integer, a contradiction.

p and q are integers, where q > 1. I am supposing that \sqrt 2 + \sqrt 3 = p/q.

 

[tazzzdo]

What are you trying to prove exactly?

If x is a rational number, p and q would have to be integers and therefore x could not be \sqrt{2} + \sqrt{3}. That's the contradiction. That means xc could never be rational for any integer c

 

[glebovg]

Suppose, beforehand we show that xc is an integer, and then show it is not. By the way, is it true that \sqrt 2 c + \sqrt 3 c is never an integer? I think it is true. Then we have arrived at a contradiction.

 

[tazzzdo]

You can use this basic proof to show the contradiction by assume xc is an integer and then proving that it's not for (\sqrt{2} + \sqrt{3})c

http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php

 

[tazzzdo]

For instance, you could do a proof of this form:

1. Assume x = \sqrt{2} + \sqrt{3} is rational.
2. Define a < xc < b, where a, b, and c are integers.
3. Prove that \sqrt{2} and \sqrt{3} are both irrational (by the contradiction proof I posted)
4. Since integers are a subset of rational numbers, and x is not in the set of rational numbers, then xc cannot be an integer for any integer c.

 

[glebovg]

Yes, I know. I just gave an example. Suppose x = \sqrt 2 + \pi. Would the same argument work? I know I left some detail out, but essentially I show that xc has to be an integer. Using the same reasoning would I arrive at a contradiction?

 

[glebovg]

For instance, you could do a proof of this form:

1. Assume x = \sqrt{2} + \sqrt{3} is rational.
2. Define a < xc < b, where a, b, and c are integers.
3. Prove that \sqrt{2} and \sqrt{3} are both irrational (by the contradiction proof I posted)
4. Since integers are a subset of rational numbers, and x is not in the set of rational numbers, then xc cannot be an integer for any integer c.

Are you sure this would work? I think the only problem is part 2. You cannot just define a < xc < b. You have to show that is the case. Correct? If I suppose x = \sqrt 2 + \pi. Would the same argument work? We know that both \sqrt 2 and \pi are irrational. If I I show that part 2 must hold, etc. I would arrive at a contradiction. Correct?

 

[tazzzdo]

I'm a little confused because it's rather vague. a and b are arbitrary, correct? You originally just assumed a < xc < b just as i did. So I don't get what you're trying to prove with that. a and b are just boundaries that imply |a/c| < x < |b/c|

 

[glebovg]

For example, let x = \sqrt 2 + \pi and assume x = p/q. Suppose we can show that xc, where c is an integer, must be an integer between two integers, namely a and b i.e. a < xc < b. If I prove that xc cannot be an integer, would it be reasonable to infer that we have a contradiction? I know it may sound confusing. I hope it makes sense. Essentially, if we show that xc must be an integer between a and b, assuming x = p/q, and then show xc cannot be an integer, will we have a contradiction (assuming we can actually show xc must be between a and b)? Can we then infer that x is irrational?

 

[tazzzdo]

You can absolutely prove that x is irrational and therefore xc is irrational and not an integer.

1. Show x is irrational by contradiction.
2. Assume a < xc < b for integers a, b, c, where xc is an integer.
3. Prove that since x is irrational, then xc is also irrational and therefore not an integer.

What's the problem? You could pick a and b to be anything outside of xc. It seems like that part is pointless. If the heart of the proof is to prove that xc cannot be an integer between some integers a and b, then just prove xc cannot be an integer (no matter what xc is, there will be integers that it lies between).

 

[glebovg]

It seems like if we let x = \sqrt 2 + \pi it does not work.