0
$\begingroup$

Prove that if $n$ is a multiple of $k$ ($n$ and $k$ are both integers), then so is $n^2$.

I know how to prove this without quantifiers, but am having difficulty working in the quantifiers. I know the problem utilizes universal specification or existential generalization, and was hoping someone could outline how I proceed with the quantifiers. I believe the symbolic statement would be: $$\forall n\ [\exists k\ (n = ck) \land (k \neq 0) \implies (n^2 = c_1k)]$$ where $c$ and $c_1$ are integers.

1 Answers 1

0

As you have written it, the statement reads: "For all $n$, there exists a $k$ such that if $k$ is not zero and $n$ is a multiple of $k$, then $n^2$ is a multiple of $k$." However, this is not what you want to prove. You want to prove the implication for all $k$ that are divisors of $n$, not merely that one such $k$ exists. The revised statement (in English) should be: "For all $n$ and $k$, if $n$ is a multiple of $k$, then $n^2$ is a multiple of $k$."

In symbols this is $$\forall\ n,k\ [(\exists\ c\!: (n = ck)) \implies (\exists\ c\!: (n^2 = ck))]$$ where all variables are integers. (Note that your original statement had the variables $c$ and $c_1$ but they were not quantified.)

  • 0
    Okay, that makes more sense. How would I go about proving that statement? – 2017-02-22
  • 0
    Hint: if $n = ck$ then $n^2 = (ck)^2 = c^2k^2 = (c^2k)k$. Are you also asking about how to write out the proof with symbols? – 2017-02-22
  • 0
    Yes- just the quantifiers. I know how to write the proof without the quantifiers (as you hinted), but the problem asks for the quantifiers to be included- I believe it has something to do with universal specialization? I was originally thinking that I had to prove that ∃ n,k~ [(∃ c:(n=ck))⟹(∃ c:(n2=ck))] was a contradiction, but I'm unsure how that could work, and figured that might just be the wrong approach. – 2017-02-22