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This is a $m + n \geq 200 \implies (m \geq 100)\lor(n \geq 100)$ argument, so my approach was to use contraposition so it is:

$(m < 100)\land(n < 100)\implies m+n<200$

Then, my unexperienced self with proofs (yet) decided to use some values and substitute for the greatest possible values for $n$ and $m$ which are supposedly $199$ if the domain is integers or $199.9999999$ if it isn't (I really was experimenting with the question as it had not assumed a domain) and said that since the sums of the two biggest values of $n$ and $m$ is in fact less than 200, then my contrapositive argument is true and therefore the main argument is also true. My proof was written in a sloppy manner because I didn't know how to formulate it but this is exactly how I solved it.

Is this correct? If it isn't, then how should it have been done? (or even if it is, I am sure there is a more systematic approach to it, I'd be happy to know what it is)

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  • $\begingroup$ If $m+n\geq 200$ then prove that $m\geq 100~~\color{red}{\text{AND}}~~n\geq 100$... What can you say about $300+50$? (You can make the statement true by changing the word "and" to something else) $\endgroup$
    – JMoravitz
    Nov 15, 2017 at 19:57
  • $\begingroup$ As for a proof for the statement $m+n\geq 200\implies m\geq 100\vee n\geq 100$, then instead of using specific values of $m$ and $n$ to try to prove $(m<100)\wedge (n<100)\implies m+n<200$, you need to be incredibly generic and use only the properties $m<100$ and $n<100$ and not use anything more specific than that about $n$ and $m$. Use the properties of addition and the less-than relation. You should know $a<b\implies a+c<b+c$. Use that property twice to get $m+n<100+n<100+100$ and then use transitivity to get $m+n<200$. $\endgroup$
    – JMoravitz
    Nov 15, 2017 at 20:01

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After you have switched around the "and"s and "or"s, the contraposed statement is easy to prove:

Assume $m<100$ and $n<100$. Then

$$ m+n < 100+n = n+100 < 100+100 = 200 $$

because $a<b \Rightarrow a+c<b+c$ which is a quite basic property of $<$.

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  • $\begingroup$ I haven't thought of this and this far more elegant than I would have ever written it. But, is my answer considered wrong? I kind of wrote it in a Discrete Maths 1 course exam so. $\endgroup$
    – Eyad H.
    Nov 15, 2017 at 20:25
  • $\begingroup$ @EyadH: Your concrete upper bounds are wrong -- if $m$ and $n$ are integers, then their sum cannot be more than $198$; if the are arbitrary real numbers, then the sum can be anything that is less than $200$, such as $199.99999995$ which is larger than the "largest value" you're proposing More seriously, you're just asserting that those are the largest possible values. At the level of detail you're working with, for a proof you're supposed to explain how you know that.. $\endgroup$
    – hmakholm left over Monica
    Nov 15, 2017 at 20:30
  • $\begingroup$ I think you misunderstood, the numbers in the questions were not the numbers specifically used or made to be the largest values. What I did in the exam paper however was just add two numbers that are very close to 100 but not quite 100 and assert that as long as both $n$ and $m$ are less than 100 each, then both will always have a sum less than 200. $\endgroup$
    – Eyad H.
    Nov 15, 2017 at 20:37
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    $\begingroup$ @EyadH. Even so, your task is to prove it for all values, not just those values close to the boundary. Of course, it sounds silly to say, but what if we took values that were much smaller than $100$ for each of $m$ and $n$, what is to stop them from "wrapping around" and somehow ending up with their sum larger than $200$? We know that this can't happen, but we have to be able to successfully explain why. Just showing it is true for a specific pair is very different than showing it is true for all pairs. $\endgroup$
    – JMoravitz
    Nov 15, 2017 at 20:54

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