1. Prove thata < b and c < 0 implies that b*c < a*c.
2. Prove that if 0 < a < b, then 0 < 1/b < 1/a
Hello, princess08!
#2 is quite simple . . .
2. Prove that: .if $\displaystyle 0 \:<\: a \:<\: b$, then: $\displaystyle 0 \:<\: \frac{1}{b} \:< \:\frac{1}{a}$
We are given: .$\displaystyle 0 \:<\: a \:< \:b$ .[1]
. . Hence, both $\displaystyle a$ and $\displaystyle b$ are positive.
Divide [1] by the positive quantity $\displaystyle ab\!:\;\;\frac{0}{ab}\:<\:\frac{a}{ab} \:<\:\frac{b}{ab} $
Therefore: .$\displaystyle 0 \:<\:\frac{1}{b} \:<\:\frac{1}{a} $
i think we can prove this by taking Shyam's proof just a little farther, and fleshing some things out.
for this proof, it is assumed you know basic properties of numbers. in particular, that the product of two negative numbers is positive.
Proof:
Assume $\displaystyle a < b$ and $\displaystyle c < 0$. Then $\displaystyle a - b < 0$. Therefore we must have $\displaystyle c(a - b) > 0$, since this is the product of two negative numbers. Thus we have $\displaystyle ac - bc > 0 \Longleftrightarrow ac > bc$, as desired.
QED