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I'm trying to prove that if $x$ is a multiple of $3$, then $x + 1$ is not a multiple of $3$. This is a rather obvious fact, but I don't think I understand the solution.

Suppose that $x$ is a multiple of $3$ but, for the sake of contradiction, that $x + 1$ is likewise a multiple of $3$. Then $x = 3k$ for some $k \in \mathbb{Z}$ and $x + 1 = 3t$ for some $t \in \mathbb{Z}$. Then $3k + 1 = 3t$, so $3k - 3t = 1$ and hence $3(k-t) = 1$. Therefore $k - t = \frac{1}{3} \not \in \mathbb{Z}$, which is a contradiction, so $x + 1$ not a multiple of $3$, as desired.

The problem here is that the deduction $k - t = \frac{1}{3}$ followed by multiplying both sides by $\frac{1}{3}$, when division is not defined in $\mathbb{Z}$. Is this allowed? I'm trying to find another way to prove this from the axioms. Another possibility is to simply argue that $3 \mid 3(k-t)$ so $3 \mid 1$, which is an absurdity, as the only divisors of $1$ are $\pm 1$, and be done at that step.

I'd appreciate some feedback on whether each of these two proof strategies are correct and which would be preferable in this instance.

JohnT
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    When you divide by 3, you are no longer in $\Bbb Z$ but in a bigger set $\Bbb Q$. It is probably cleaner to say that $3(k-t)=1$ implies that $3$ divides $1$, a contradiction: the only divisors of $1$ are $1,-1$ – markvs Jan 12 '22 at 04:03
  • @markvs Am I allowed to do that, though, when I'm working in $\mathbb{Z}$? I'm in effect using the properties of $\mathbb{Q}$ to deduce (lack of equality) in $\mathbb{Z}$. – JohnT Jan 12 '22 at 04:06
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    Since $\Bbb Z$ is inside $\Bbb Q$ then "working" in the smaller set implies "working" in the bigger set, but I have explained how to avoid it altogether. – markvs Jan 12 '22 at 04:07
  • @markvs How do you prove that the only divisors of $1$ are $1$ and $-1$ without "working" in $\mathbb{Q}$? The proof I've seen relies on showing that if $|d|>1$, then $\left|\frac{1}{d}\right|<1$ and hence cannot be a divisor of $1$. – Alan Abraham Jan 12 '22 at 04:40
  • Let $1=mn$ for some integers $m,n\ne 0$. Then if $m$ or $n$ is negative, replace it by its negative and $1$ by $-1$. So you get $\pm 1=mn$ where $m,n>0$. Then the LHS is $1$ (because the RHS is positive). Then $mn\ge m$ and $mn\ge n$. Hence $1\ge m, n$, so the integers $m,n$ are $1$. As you can see, rational numbers are not needed. – markvs Jan 12 '22 at 05:15
  • The fact that 1 is not a multiple of 3 is a fact that can be proven without any reference whatsoever to the field of rational numbers. It is a fact about integers, plain and simple. – Jake Mirra Jan 12 '22 at 05:32
  • Both are correct. Since $,a\mid b,$ in $\Bbb Z\iff \frac{b}a\in \Bbb Z$ one can use fractions to (dis)prove divisibilities, and in more complex problems fractional arithmetic may prove simpler, e.g. see here for an interesting important example. More generally Consecutive Numbers are coprime, and the proofs there may prove illuminating. – Bill Dubuque Jan 12 '22 at 08:48

2 Answers2

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Suppose that $ x $ is a multiple of $3 $ but, for the sake of contradiction, that $x+1$ is likewise a multiple of $3$ . Then $x=3k $ for some $ k\in \Bbb{Z }$ and $ x+1=3t $ for some $t\in \Bbb{Z}.$ Then $3k+1=3t$ ,

$3k−3t=1 $ and hence $ 3(k−t)=1$

As, $k, t\in \Bbb{Z}\implies s=k-t\in\Bbb{Z}$

And $3s=1$.

Implies $3|1$ , which is absurd.

Sourav Ghosh
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What is done by the proof is the following:

  1. First, the set of integers is closed under addition and subtraction. So as $k$ and $t$ are both integers, then so must $k-t$ be an integer.

  2. Algebra was used to derive the equation $3(k-t)=1$, and thus $k-t = 1/3$. If an equation is true in a field [here $\mathbb{Q}$], then it remains true if you multiply both sides by a nonzero element in the field [here $\frac{1}{3}$]. ETA: Or if you prefer, that the equation $3(k-t)=1$ holds and $3$ does not have a multiplicative inverse in $\mathbb{Z}$, gives $(k-t)$ cannot be an integer after all.

  3. Putting 1. and 2. together yields the desired contradiction as both cannot be simultaneously true.

I will elaborate on Step 2. For the claim to be true in the first place, what is actually needed is that $3$ does not have a multiplicative inverse in $\mathbb{Z}$; put another way, there is no other integer $a$ in $\mathbb{Z}$ such that $3a$ is $1$. In fact this result would not hold if we were working in a ring where $3$ has a multiplicative inverse; these exist, an example is $\mathbb{Z}/M\mathbb{Z}$ for any integer $M$ satisfying $\gcd(M,3)=1$.

Mike
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    The OP understands the proof which you copied. The question is why the division by $3$ is possible when working in $\Bbb Z$. – markvs Jan 12 '22 at 04:10
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    I'm curious if this is another way to say the following. Treat $k-t$ as a variable (say, real). Then $3(k-t) = 1$ has a unique solution in $\mathbb{R}$ and hence a unique solution in $\mathbb{Z}$ if $k-t \in \mathbb{Z}$; otherwise, it has no solution. We found a solution in $\mathbb{R}$ and in $\mathbb{Q}$, but it's not an integer. If there were some integer solution, the solution in $\mathbb{R}$ would not be unique, so we have a contradiction. – JohnT Jan 12 '22 at 04:12
  • I'm probably overly complicating it quite a bit, but is the logic of what I said above sound? – JohnT Jan 12 '22 at 04:12
  • @markvs I saw that and editted. I am not sure what is meant precisely by "working in $\mathbb{Z}$". In particular, there is no reason why one cannot multiply both sides of an equation by a nonzero rational. – Mike Jan 12 '22 at 04:17
  • @JohnT or you could use the fact that if an equation is true in a field, such as $\mathbb{Q}$, then the equation is still true after multiplying by a nonzero element in the field, here $\frac{1}{3}$. – Mike Jan 12 '22 at 04:24
  • @Mike: When you divide by $3$, you are working in $\Bbb Q$. – markvs Jan 12 '22 at 04:39
  • The key for the result to hold though, is that $3$ does not have a multiplicative inverse in $\mathbb{Z}$. If we were to replace $\mathbb{Z}$ with a ring where $3$ does have a multiplcative inverse then the result would not hold. – Mike Jan 12 '22 at 04:45
  • Meanwhile, what is the problem, mathematically, with "working in $\mathbb{Q}$" in the first place? – Mike Jan 12 '22 at 04:53