What proof technique can be used to solve this problem involving odd numbers?

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Discussion Overview

The discussion revolves around the proof technique applicable to a problem involving odd numbers, specifically the statement: "For all x in Z, if 2^(2x) is odd, then 2^(-2x) is odd." Participants explore whether this can be approached as a direct proof or if it falls under vacuous proofs, given the nature of the premise.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that since 2^(2x) is always even for integer x, the statement can be shown vacuously, questioning the classification of the problem under direct proofs.
  • Another participant raises a concern about the interpretation of 2^(-2x) when x is positive, asking how even/odd applies to rational numbers.
  • There is a suggestion that the problem might contain a typographical error, questioning if the exponent should be -22x instead of -2x.
  • One participant draws a parallel to a similar logical structure, stating that if the premise is always false, the implication is vacuously true, and wonders if this reasoning applies here.
  • Another participant asserts that the proof can be direct since the conclusion follows from the premise being false.
  • Several participants express uncertainty about the validity of the statement for values other than x=0, with one noting that 2^(2x) can indeed be an odd integer.
  • There is a discussion about the phrasing of the problem, with one participant clarifying that the original statement is valid and that it can be approached as a direct proof by showing that the only integer making 2^(2x) odd is zero.
  • One participant acknowledges a misunderstanding regarding the phrasing of the problem and admits to missing the validity of the original statement.

Areas of Agreement / Disagreement

Participants express differing views on whether the problem can be classified as a direct proof or if it should be considered vacuous. There is no consensus on the correct approach, and multiple interpretations of the problem persist.

Contextual Notes

Participants highlight potential ambiguities in the problem's phrasing and the implications of the premise being false. The discussion reflects uncertainty regarding the application of proof techniques and the nature of the statements involved.

medwatt
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Hello,
Found a book on proofs and went to the exercise section. The proofs are fairly easy. The problem is the question I tried is listed under "direct proofs". I wasn't able to use direct proofs. Here's the question:
For all x in Z, if 2^(2x) is odd, then 2^(-2x) is odd.

My thoughts:
I thought this was a very easy problem because all I had to do was show that 2^(2x)=4^x which is always even and so the proof follows vacuously.
So I wonder why the author listed this problem under "direct proofs" and not under "vacuous proofs" which also has a section of its own. Is there a direct proof ? Is my proof wrong?

Thanks
 
Last edited:
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medwatt said:
Hello,
Found a book on proofs and went to the exercise section. The proofs are fairly easy. The problem is the question I tried is listed under "direct proofs". I wasn't able to use direct proofs. Here's the question:
For all x in Z, if 2^(2x) is odd, then 2^(-2x) is odd.

If x > 0, shouldn't 2^(-2x) be equal to 1/[2^(2x)]? How does even/odd work for a rational, non-integer number?
 
medwatt said:
For all x in Z, if 2^(2x) is odd, then 2^(-2x) is odd.

Perhaps it should be -22x? Are you sure that's what's written?
 
Yes I'm sure. Here's a pic of the question:
2q1s6mv.png


Why are you insisting on 2^(-2x) if you the premise 2^(2x) is always false ?? I mean the question seems to be similar to the statement: if x^2+1<0, then 1 is even. Since x^2+1<0 is always false, then the statement if x^2+1<0, then 1 is even is always true. I hope I'm explaining myself.
What I'm asking is can I use the same train of thought to prove the result as shown in the picture. I proved it using a vacuous proof. The question is listed under "direct proofs" meaning that I have to show that for all premises the conclusion has to be true.
 
Last edited:
You showed directly that the statement is always true, therefore it is a direct proof. I mean, you didn't say "assume the statement is false, then...".
 
the only way i see this working is if x=0, unless I am missing something
 
cragar said:
the only way i see this working is if x=0, unless I am missing something

This is a good point, ##2^{2x}## can be an odd integer. I didn't look at the question closely at all because Medwatt was asking about whether a proof like he suggested would be a direct proof.
 
medwatt said:
Hello,
Found a book on proofs and went to the exercise section. The proofs are fairly easy. The problem is the question I tried is listed under "direct proofs". I wasn't able to use direct proofs. Here's the question:
For all x in Z, if 2^(2x) is odd, then 2^(-2x) is odd.

My thoughts:
I thought this was a very easy problem because all I had to do was show that 2^(2x)=4^x which is always even and so the proof follows vacuously.
So I wonder why the author listed this problem under "direct proofs" and not under "vacuous proofs" which also has a section of its own. Is there a direct proof ? Is my proof wrong?

Thanks

x=0, so it isn't vacuous.
 
medwatt said:
For all x in Z, if 2^(2x) is odd, then 2^(-2x) is odd.
pwsnafu said:
Are you sure that's what's written?
medwatt said:
Yes I'm sure. Here's a pic of the question:
2q1s6mv.png
"For all x in Z..." is NOT what is in the question according to your picture of it. This means something entirely different.
 
  • #10
skiller said:
"For all x in Z..." is NOT what is in the question according to your picture of it. This means something entirely different.

Basically this.

The direct proof would start by proving the only integer that makes 22x odd is zero (easy enough), then substituting it into 2-2x to get 1 which is odd.
 
  • #11
pwsnafu said:
Basically this.

The direct proof would start by proving the only integer that makes 22x odd is zero (easy enough), then substituting it into 2-2x to get 1 which is odd.
Actually, yet again, I'm an idiot!

"For all x in Z, prove that if..." is equally as valid as "Let x be in Z. If..."
 
  • #12
skiller said:
Actually, yet again, I'm an idiot!

"For all x in Z, prove that if..." is equally as valid as "Let x be in Z. If..."

Huh, you're right, totally missed that.
 

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