1

I've been working on a proof and this issue came up. Let's say I've made a claim that $ 7 \nmid 6x^2 + 13x - 5 $. It follows then that $ 7 \nmid (3x -1)(2x+5) $. But does it follow that $ 7 \nmid (3x-1) \wedge 7 \nmid (2x + 5) $?

& also I have an additional question which kind of relates to the above. I know that for any $5$ sequences of numbers in the domain $ \forall n \in \mathbb Z^+ $ there's at least one number in that sequence that is divisible by $5$.

i.e.

$ P = \{n, n+1, n+2, n+3, n+4\} $ $ n \in \mathbb Z^+$

$ \exists g \in P, 5 \mid g$

Now to me this is something that's self-evident and doesn't need explaining. However the people I'm writing this proof for are not mathematically inclined and may not be able to follow along.

Is there some kind of mathematical property that can allow me to write a more elegant solution? I'm very new to writing proofs.

tuba09
  • 323
  • For all $a,,b,,c \in \mathbb{Z}$, $a \mid b \Rightarrow a \mid bc$ and $a \mid c \Rightarrow a \mid bc$. So the answer to the first question is yes. – Daniel Fischer Jun 27 '13 at 14:18

1 Answers1

1

In general, if $a|b$ then $a|bc$ for all $c$. So if $a\not\mid bc$ then $a\not\mid b\land a\not\mid c$.

The second part follows from the division algorithm - if $a,b$ are real numbers and $b\neq 0$ then we can find a $q,r$ such that $a=bq+r$ and $0\leq r<b$. It isn't clear that is more helpful to people not inclined towards mathematics.

The other approach is to use induction. If one of $n,n+1,n+2,n+3,n+4$ is divisible by $5$, can you show that one of $n+1,n+2,n+3,n+4,n+5$ is divisible by $5$?

Thomas Andrews
  • 177,126