Prooving a statement with a Lemma

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The discussion centers on proving that if two integers divide each other, they must be equal, using the lemma that states if the product of two integers is 1, then both integers equal 1. Initial confusion arises regarding the lemma's validity, particularly with examples involving fractions, which are clarified as not applicable since they are not integers. The proof process involves understanding the definitions of divisibility and manipulating equations to show that if one integer divides another, the multipliers must equal 1. Ultimately, participants confirm that through proper assumptions and algebraic manipulation, it can be shown that the integers are indeed equal. The conversation concludes with a successful resolution of the proof.
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Homework Statement


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.


Homework Equations





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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alexk307 said:
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?
 
alexk307 said:

Homework Statement


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.


Homework Equations





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.
 
Mark44 said:
3/4 and 4/3 aren't integers, nor are 2/3 and 3/2.

Oh man, that's embarrassing... Thanks
 
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?
 
alexk307 said:
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?
 
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.
 
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.
 
alexk307 said:
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?
 
  • #10
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.
 
  • #11
alexk307 said:
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.
 
  • #12
very helpful, thanks for the walkthrough!
 
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