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Prooving a statement with a Lemma

by alexk307
Tags: discrete, math, number theory, proof
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alexk307
#1
Mar29-11, 02:32 PM
P: 27
1. The problem statement, all variables and given/known data
Using the lemma below, prove that if two integers divide each other, then they are equal

Lemma: If the product of two integers is 1, then the integers each equal 1.


2. Relevant equations



3. The attempt at a solution
Very lost here, I can format the proof but I don't know where to start it. Also, isn't the lemma false? If a*b = 1, then a and b equal 1. What about 3/4 * 4/3 = 1? or 2/3 * 3/2 = 1?
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gb7nash
#2
Mar29-11, 04:18 PM
HW Helper
P: 805
Quote Quote by alexk307 View Post
Also, isn't the lemma false? If a*b = 1, then a and b equal 1. What about 3/4 * 4/3 = 1? or 2/3 * 3/2 = 1?
Ah, but you're forgetting the hypothesis of the lemma. What's the hypothesis? The hypothesis is assumed to be true so....

Once you understand the lemma, what does it mean for one integer to divide another integer?
Mark44
#3
Mar29-11, 04:24 PM
Mentor
P: 21,249
Quote Quote by alexk307 View Post
1. The problem statement, all variables and given/known data
Using the lemma below, prove that if two integers divide each other, then they are equal

Lemma: If the product of two integers is 1, then the integers each equal 1.


2. Relevant equations



3. The attempt at a solution
Very lost here, I can format the proof but I don't know where to start it. Also, isn't the lemma false? If a*b = 1, then a and b equal 1. What about 3/4 * 4/3 = 1? or 2/3 * 3/2 = 1?
3/4 and 4/3 aren't integers, nor are 2/3 and 3/2.

alexk307
#4
Mar29-11, 05:43 PM
P: 27
Prooving a statement with a Lemma

Quote Quote by Mark44 View Post
3/4 and 4/3 aren't integers, nor are 2/3 and 3/2.
Oh man, that's embarrassing... Thanks
alexk307
#5
Mar29-11, 06:09 PM
P: 27
okay so,
I assumed that a and b are both integers and that a/b = 1.
Should I also assume that c*d=1 according to the lemma that if cd=1 then c and d are 1?
gb7nash
#6
Mar29-11, 06:12 PM
HW Helper
P: 805
Quote Quote by alexk307 View Post
okay so,
I assumed that a and b are both integers and that a/b = 1.
Should I also assume that c*d=1 according to the lemma that if cd=1 then c and d are 1?
Start with the hypothesis of what you're trying to prove. Using math terms, what does it mean if a divides b? b divides a?
alexk307
#7
Mar29-11, 07:18 PM
P: 27
Okay, so if a divides b, then a = bq, where q is the multiplier, also there is no remainder.

so I see that a = bq and b=aq

solving for q... q^2=1 and q=1

I don't know if I'm onto anything by saying q*q=1, and the lemma are related.
alexk307
#8
Mar29-11, 07:23 PM
P: 27
just got it I think. because q*q = 1, q must equal 1. Therefor a = bq and b = aq can be reduced to a=b and b=a.
gb7nash
#9
Mar29-11, 07:46 PM
HW Helper
P: 805
Quote Quote by alexk307 View Post
so I see that a = bq and b=aq
Not quite. It's true that a = bq, where q is an integer. However, you cannot say b = aq. It could be a different integer, so the simple solution is to make b = ar, where r is an integer.

So now that you have the two equations:

a = bq
b = ar

What can we do?
alexk307
#10
Mar29-11, 07:57 PM
P: 27
oh okay so a=bq and b=ar solving for then rq=1.

So I had the right idea, but I can't say that they're both able to be divided by the same integer, but that rq=1

and then r and q both equal 1, so a=b and b=a.
gb7nash
#11
Mar29-11, 08:05 PM
HW Helper
P: 805
Quote Quote by alexk307 View Post
oh okay so a=bq and b=ar solving for then rq=1.

So I had the right idea, but I can't say that they're both able to be divided by the same integer, but that rq=1

and then r and q both equal 1, so a=b and b=a.
You got it.
alexk307
#12
Mar29-11, 08:25 PM
P: 27
very helpful, thanks for the walkthrough!


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