# If $a<b$, $c<d$ then $a+c<b+d$?

This is how I managed to prove it:

I know $a \lt b$ and $c \lt d$ thus, $b-a$ and $d-c$ are real positive numbers. Then, $b-a + d-c \gt 0$ and because of this $a+c \lt b+d$

Did I prove it right?

• It seems correct to me! Commented Aug 20, 2017 at 22:05
• You are using that $x \gt y \iff x - y \gt 0\,$, and $x \gt 0, y \gt 0 \implies x+y \gt 0\,$. Is that something you proved before, or are otherwise allowed to use?
– dxiv
Commented Aug 20, 2017 at 22:07
• I proved that before. But im going to include it in the proof. Thank you. Commented Aug 20, 2017 at 22:09
• @TheNicouU If you happen to have also proved before that $x \gt y \implies x+z \gt y+z\,$, then you can just use that twice: $a+c \lt b+c \lt b+d$.
– dxiv
Commented Aug 20, 2017 at 22:12
• I'd just try something along the lines of if a<b than b=xa for some value x>1, if c<d then d=yc for some value y>1 so we can rewrite the other inequality as a+c<xa+yc and with x,y >1 ...
– user451844
Commented Aug 20, 2017 at 22:14

your way is correct,another one is add both side $c$ to the $a<b$ so that $$a+c<b+c\\$$ then add $b$ to the $c<d$ so that $$c+b<d+b$$ finally we can conclude that $$a+c<b+d$$

Yes. Formalizing, you might write something like:

$$(a<b)\land(c<d)$$ $\implies$ (using $x<y\iff x+z<y+z$) $$(0<b-a)\land(0<d-c)$$ $\implies$ (using $x<y\iff x+z<y+z$) $$(d-c<b-a+(d-c))\land(0<d-c)$$ $\implies$ (transitivity) $$0<b-a+(d-c)$$ $\implies$ (using $x<y\iff x+z<y+z$ and associativity of addition) $$a+c<b+d$$

• But that is not the axiom system OP is using. The OPs proof is assuming the existence of a set of "positive" values, with certain properties, and defines $a<b$ as "$b-a$ is a positive value." Commented Aug 20, 2017 at 22:13
• @ThomasAndrews I don't quite understand. How do you know this isn't the axiom system the OP is using? It's not stated in the question. Commented Aug 20, 2017 at 22:25
• @ThomasAndrews To me, '$b−a$ is a positive value' $\equiv$ '$b-a>0$'. ie. $a<b\iff 0<b-a$ Commented Aug 20, 2017 at 22:34

I would like to give a sweet and simple proof:

• $a$ < $b$ ---------------------(1)

• $c$ < $d$ ---------------------(2)

• now simply add the two equations:

• $a+c$ < $b+d$

• hope it helps you!

• This is maybe the intuition behind the OP's statement, but it is nowhere near a proper proof. Commented Aug 26, 2018 at 22:03