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

[medwatt]

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

 

[SteamKing]

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?

 

[pwsnafu]

For all x in Z, if 2^(2x) is odd, then 2^(-2x) is odd.

Perhaps it should be -2^2x? Are you sure that's what's written?

 

[medwatt]

Yes I'm sure. Here's a pic of the question:

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.

 

[verty]

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...".

 

[cragar]

the only way i see this working is if x=0, unless I am missing something

 

[verty]

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.

 

[Hornbein]

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.

 

[skiller]

For all x in Z, if 2^(2x) is odd, then 2^(-2x) is odd.

Are you sure that's what's written?

Yes I'm sure. Here's a pic of the question:

"For all x in Z..." is NOT what is in the question according to your picture of it. This means something entirely different.

 

[pwsnafu]

"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 2^2x odd is zero (easy enough), then substituting it into 2^-2x to get 1 which is odd.

 

[skiller]

Basically this.

The direct proof would start by proving the only integer that makes 2^2x 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..."

 

[pwsnafu]

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.