If $a\lt{b}$ and $c\le{d}$, prove that $a+c\lt b+d$.

This seems like a basic proof and I think this is how it goes: $$c \le d, \text{ Given }$$ $$a+c \le a+d$$ $$a+c \lt b+d, \text{ since } a \lt b$$

Is that all I need? I'm thinking this does it all quickly and concisely, but I have had trouble with proofs in my classes.

  • 3
    $\begingroup$ Yes, or just go $a < b \implies a+c<b+c$, so since $c \leq d \implies b + c \leq b + d$, we have $a+c < b+c \leq b+d$, i.e. $a+c < b+d$. $\endgroup$ – Ryker Feb 17 '14 at 20:10
  • $\begingroup$ Is $(b+d)-(a+c)>0$? $\endgroup$ – David Mitra Feb 17 '14 at 20:10
  • $\begingroup$ I'd do it $(a-b)+(c-d)\lt 0$ but it's all the same in the end $\endgroup$ – Mark Bennet Feb 17 '14 at 20:18
  • $\begingroup$ What is the reason you did it your way, @Ryker, versus my way? It is just semantical? I've noticed that the things posted have a middle argument between a less than and a less than or equal to sign. Is that preferred when proving to see the logical reason? $\endgroup$ – Faffi-doo Feb 17 '14 at 20:25
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    $\begingroup$ @Faffi-doo , guys, wow! :) Perhaps I should have added a smiley in the end of the comment to avoid misunderstandings.. 1. You don't need to be so apologetic! :) Choose what is most correct for you. Always, and not just in Math. 2. I will not hold anything against you even if you actually did not like my answer. $\endgroup$ – user76568 Feb 17 '14 at 21:21

We have that for any $c$: $$a<b \implies a+c<b+c$$ And for any $b$: $$c\leq d \implies b+c \leq b+d$$.

Hence: $a+c < b+c \leq b+d$, that is: $$a+c < b+d$$.


$$(a < b) \land (c \leq d) \iff a-b < 0 \leq d - c \implies a-b<d-c \iff a+c < b + d$$


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