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Here is the proof problem:

Prove that:

$$\forall a \in \mathbb{N} \ \ \forall p \in \mathbb{N}\setminus \{0\} \ \exists x \in \left\{ a+k \ | \ k \in 0.. (p-1)\right\}(p|x)$$

First, I need some help translating this. I think I've done it correctly. $x$ would belong to the set cardinality $1$ of $a+k$, which would mean x is either the one element, or the empty set, which doesn't work.

Then, I know that this isnt true of all a and p, plenty of counterexamples.

So, what strategy would I use to tackle this nonexistence proof?

Bill Dubuque
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Bean Greasy
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    Please don't use pictures. – Dietrich Burde Mar 07 '21 at 19:24
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    Please use MathJax to typeset your question, rather than relying on images; images are not searchable and generally are not accessible; and in this case, people need to navigate away from your post to see it. Here’s a MathJax Tutorial – Arturo Magidin Mar 07 '21 at 19:24
  • What set of cardinality $1$? The given set is of cardinality $p$. What non-existence proof? You are asked to show the existence of such an $x$. – saulspatz Mar 07 '21 at 19:37
  • What you’re being asked to prove is that if $p$ is any positive integer, every set of $p$ consecutive non-negative integers contains a multiple of $p$. – Brian M. Scott Mar 07 '21 at 20:13
  • @saulspatz but it's not true – Bean Greasy Mar 07 '21 at 20:25
  • @BrianM.Scott which is not a true statement, right? – Bean Greasy Mar 07 '21 at 20:26
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    @BeanGreasy: It is a true statement. One of the integers $a+k$ (for $k\in{0,\ldots,p-1}$) is congruent to $0$ modulo $p$. – Brian M. Scott Mar 07 '21 at 20:27
  • It is a true statement. I was trying to point out that you seem to misunderstand what is being asked. Brian's restatement is correct. I sympathize with you. It always seems to me that writing statements like this, instead of using words wherever possible, just obscures what is being said. Of course, maybe the intent of the exercise, or at least part of it, is to practice decoding stuff like this. – saulspatz Mar 07 '21 at 20:32
  • @BrianM.Scott Ok I didn't fully construe the implications of k not included under the universal quantifier. it seems to me that if a is 0, then k can be 0 so it's true. and if a is non-0, then k can be p-a, so 2p, which makes it true. But how can i show this in a proof? – Bean Greasy Mar 07 '21 at 21:24
  • @BeanGreasy: I’ve posted an answer that contains a fairly big hint for how to do that. – Brian M. Scott Mar 07 '21 at 21:29

1 Answers1

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HINT: By the division algorithm there are non-negative integers $q$ and $r$ such that $a=pq+r$, and $0\le r\le p-1$. Use this $r$ to find a $k\in\{0,\ldots,p-1\}$ such that $a+k$ is a multiple of $p$.

Brian M. Scott
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  • so what proof strategy would be best for validating this existence? – Bean Greasy Mar 07 '21 at 23:22
  • @BeanGreasy: There is a very simple formula for finding $k$ from $r$. What must $k$ be if $pq+r+k$ is to be a multiple of $p$, and $k$ is to be in the set ${0,\ldots,p-1}$? – Brian M. Scott Mar 07 '21 at 23:25
  • I dont know if this is a formula, but

    if a < p, k is p-a;

    if a>p, k= a-p

     – Bean Greasy Mar 07 '21 at 23:26
  • @BeanGreasy: If $a=3p+1$, say, then $k$ is obviously not $a-p$, since $a-p$ is too big. Forget $a$. Look at $r$. And perhaps consider some numerical examples with $p=5$, say, and $a$ bigger than $10$. – Brian M. Scott Mar 07 '21 at 23:28
  • if r= -k, then a-r=pq, which suggests that a+k =pq which suggests it's a multiple – Bean Greasy Mar 08 '21 at 02:24
  • @BeanGreasy: If your $x$ here is the one in the problem, that’s incorrect; if it’s anything else, it’s currently undefined. I can guarantee, however, that the $k$ that you want is in general not going to be of the form $a-xp$. – Brian M. Scott Mar 08 '21 at 02:35
  • I've since edited it – Bean Greasy Mar 08 '21 at 03:07