Irrational numbers

  • Thread starter glebovg
  • Start date
  • #1
164
1
Can anyone explain what is wrong with my reasoning?

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

p and q are integers, where q > 1. I am supposing that [itex]\sqrt 2 + \sqrt 3 = p/q[/itex].
 
Last edited:

Answers and Replies

  • #2
47
0
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 [itex]\sqrt{2}[/itex] + [itex]\sqrt{3}[/itex]. That's the contradiction. That means xc could never be rational for any integer c
 
  • #3
164
1
Suppose, beforehand we show that xc is an integer, and then show it is not. By the way, is it true that [itex]\sqrt 2 c + \sqrt 3 c[/itex] is never an integer? I think it is true. Then we have arrived at a contradiction.
 
  • #5
47
0
For instance, you could do a proof of this form:

1. Assume x = [itex]\sqrt{2}[/itex] + [itex]\sqrt{3}[/itex] is rational.
2. Define a < xc < b, where a, b, and c are integers.
3. Prove that [itex]\sqrt{2}[/itex] and [itex]\sqrt{3}[/itex] 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.
 
  • #6
164
1
Yes, I know. I just gave an example. Suppose [itex]x = \sqrt 2 + \pi[/itex]. 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?
 
Last edited:
  • #7
164
1
For instance, you could do a proof of this form:

1. Assume x = [itex]\sqrt{2}[/itex] + [itex]\sqrt{3}[/itex] is rational.
2. Define a < xc < b, where a, b, and c are integers.
3. Prove that [itex]\sqrt{2}[/itex] and [itex]\sqrt{3}[/itex] 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 [itex]x = \sqrt 2 + \pi[/itex]. Would the same argument work? We know that both [itex]\sqrt 2[/itex] and [itex]\pi[/itex] are irrational. If I I show that part 2 must hold, etc. I would arrive at a contradiction. Correct?
 
Last edited:
  • #8
47
0
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|
 
  • #9
164
1
For example, let [itex]x = \sqrt 2 + \pi[/itex] and assume [itex]x = p/q[/itex]. 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?
 
Last edited:
  • #10
47
0
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).
 
  • #11
164
1
It seems like if we let [itex]x = \sqrt 2 + \pi[/itex] it does not work.
 

Related Threads on Irrational numbers

  • Last Post
Replies
15
Views
4K
  • Last Post
Replies
8
Views
2K
Replies
26
Views
17K
  • Last Post
Replies
18
Views
12K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
958
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
2
Views
3K
Replies
2
Views
1K
Top