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Prove or disprove: If a, b belong to the set of positive integers, and if a divides b and b divides a, then a=b. Does this hold if if a,b are not necessarily positive? Why or Why not?

Here is what I have:

If a and b are integers, we say a divides b if there exists an integer k so that b=ak. Thus: $$b=ak$$ $$a=bj$$

$$a=akj$$ $$kj=1$$

Since a divides b and b divides a, we know that k,j must both be integer, which means k and j must both equal 1 or -1. This means:

$$a=b$$

This proves that a=b when we assume that a,b are both positive integers.

For the second part of the question, I am asked if I can still say a=b if the assumption about a, b is relaxed so that a,b belong to the integers (not the positive integers.) I think The answer is no, I cannot say that a=b for every case when we allow a,b to also be negative integers.

Here is my reasoning (I use the same idea that a=bj and b=ak here)

$$+/-a=akj$$ $$+/-1=kj$$

So $ |a|=|b|$ but not $a=b$

I'm very new to proofs so any suggestions or thoughts are greatly appreciated.

123
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  • Right $|a|=|b|$, but remember your assumption about $a,b$? – Thomas Andrews Sep 21 '14 at 21:14
  • For the first part we assume a,b belong to positive integers. For the second part we relax that assumption to a,b belong to the integers. Or, do you mean the assumption that a divides b and b divides a should resolve this? – 123 Sep 21 '14 at 21:15
  • In the case that negative numbers are allowed you are nearly there: you are not looking for a proof, but for a counterexample (one will do). Just make sure you don't choose $a=0$. – Mark Bennet Sep 21 '14 at 21:18
  • It's really unclear that what you are calling "the second part" is not just your expression of the proof of the first part. – Thomas Andrews Sep 21 '14 at 21:18
  • For example, the sentence, " think I have this but I want to make sure before submitting this for evaluation..." seems to indicate that the rest is your proof of the first part. – Thomas Andrews Sep 21 '14 at 21:19
  • Ah! Okay, I will do some editing to clear that up. Thanks! – 123 Sep 21 '14 at 21:21

2 Answers2

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Since $a,b \in \mathbb{Z^+}$, $a=b$ most definitely, as you have proved. You don't have to worry about negatives here if it is given that $a,b$ are positive integers. If $a,b \in \mathbb{Z}$, then a counterexample would be $a=−6,b=6$. Hence, unless $a,b$ are strictly positive, the statement is not true.

user30625
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  • Oh yes...so simple. I will just present a counter example for the second part! For some reason I was thinking I needed to do it in a more general way. – 123 Sep 21 '14 at 21:28
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Really, if $a/ b & b/a$ then we know $a=b$ because if the values of $a$ and $b$ are same then they will divide completely each other so, Their values will be either positive or negative.

So, $a=\pm b$.

Hence proved

Nosrati
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Sonaton Ramchiary
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