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

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In summary, the author listed a problem under direct proofs, but it's actually a vacuous proof. The author is asking for someone to show that for all integers 0-2^x, 2^(-2x) is even, but this is easily proven using the identity 2^x=-2^x.
  • #1
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
 
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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
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  • #5
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...".
 
  • #6
the only way i see this working is if x=0, unless I am missing something
 
  • #7
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.
 
  • #8
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.
 
  • #9
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.
 

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

What is a proof technique?

A proof technique is a method used in mathematics and other sciences to prove the validity of a statement or theorem. It involves a logical sequence of steps that demonstrate the truth of a claim.

How do you choose which proof technique to use?

Choosing the right proof technique depends on the nature of the statement or theorem being proved. Some common techniques include direct proof, proof by contradiction, and proof by induction. The choice also depends on the personal preference and experience of the mathematician or scientist.

What is the difference between a direct proof and an indirect proof?

In a direct proof, the statement or theorem is proved directly using logical steps and previously established facts. In contrast, an indirect proof, also known as proof by contradiction, assumes the opposite of the statement to be true and then shows that it leads to a contradiction, thus proving the original statement to be true.

When should I use proof by induction?

Proof by induction is a technique used to prove statements that hold for all natural numbers. It is particularly useful when the statement involves a recursive definition, where the truth of the statement for one value depends on its truth for the previous value.

Are there any other proof techniques besides direct proof, proof by contradiction, and proof by induction?

Yes, there are several other proof techniques, such as proof by construction, proof by exhaustion, and proof by counterexample. Each technique has its own strengths and weaknesses, and the choice of technique depends on the nature of the statement being proved.

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