Prove:
If 0 < a < b and 0 < c < d, then ac < bd
I know that I need to use the order axioms. However, the order axioms I have are not helping me. I am assuming that I need to be using some type of corrollary to one of the order axioms...
Prove:
If 0 < a < b and 0 < c < d, then ac < bd
I know that I need to use the order axioms. However, the order axioms I have are not helping me. I am assuming that I need to be using some type of corrollary to one of the order axioms...
Start with the statement on the left and multiply it by c.
we know this is okay because c is > 0
$\displaystyle c\cdot 0 < c \cdot a < c \cdot b \implies 0 < ac <bc$
Now multiply the second by b to get
$\displaystyle 0 < bc<bd$
Now chaining the two together we get
$\displaystyle 0 < ac < bc < bd \implies ac < bd$