Prove if xy < 140 then x < 10 or y< 14 by Contraposition

Question

I have the following proof for this statement -

Assume $x$ and $y$ are both integers. Prove by contraposition, if $xy < 140$ then $x < 10$ or $y< 14$.

Assume that $x ⩾ 10$ and $y ⩾ 14$, then $xy ⩾ 10\cdot14 = 140$

My attempt at a proof is as above, however my answer seems rather short, is this enough to be a valid proof? Is there any additions I should make?

+1 I regard your proof as valid. – user2661923 Oct 04 '20 at 13:42 — user2661923, Oct 04 '20 at 13:42
You seem to be using $x,y$ and $m,n$ interchangeably. – lulu Oct 04 '20 at 13:46 — lulu, Oct 04 '20 at 13:46
I understand now. I'll upvote. – cosmo5 Oct 04 '20 at 13:48 — cosmo5, Oct 04 '20 at 13:48

score 1 · Answer 1 · answered Oct 04 '20 at 13:26

1

The contraposition of if xy < 140 then x < 10 or y< 14 is :

if not(x < 10 or y< 14) then not(xy<140)

which is the same than

if (x >= 10 and y >= 14) then (xy >= 140)

(which is what you stated)

This proposition is true because both of the inequalities concern real positive numbers

There's no need in adding anything else.

answered Oct 04 '20 at 13:26

Arnaud Feldmann

You can also see it better by absurd : if (xy < 140 and 10 <= x) then xy < 14x and you can divide by x because its positive and not equal to zero. Hence y <14. – Arnaud Feldmann Oct 04 '20 at 13:36
+1 Long way round, but still good detail. No explanation why anyone downvoted. Reminded of Star Trek (TOS) episode at the OK corral: (Spock): when the physical laws cease to exist, reality must be questioned. What was the downvoter looking for, graphics? – user2661923 Oct 04 '20 at 13:46

score 0 · Answer 2 · answered Oct 04 '20 at 14:09

0

Yes, the argument is that simple.

You just needed to fix the typos in your question using $x$ and $y$ throughout (or $m$ and $n$). I did it for you.

answered Oct 04 '20 at 14:09

egreg

2 Answers2