1. ## prove inequality property

1. Prove that
a < b and c < 0 implies that b*c < a*c.

2. Prove that if 0
< a < b, then 0 < 1/b < 1/a

2. 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: .$\displaystyle 0 \:<\:\frac{1}{b} \:<\:\frac{1}{a}$

3. Originally Posted by princess08
1. Prove that
a < b and c < 0 implies that b*c < a*c.

2. Prove that if 0
< a < b, then 0 < 1/b < 1/a

Since, a < b

Multiply both sides with a negative number 'c' (because c<0)

ac > bc (multiplying with a negative number reverses the inequality sign to >)

so, bc < ac

4. Originally Posted by Shyam
Since, a < b

Multiply both sides with a negative number 'c' (because c<0)

ac > bc (multiplying with a negative number reverses the inequality sign to >)

so, bc < ac
i think what the problem is getting at is why do we reverse the inequality sign? i think they want us to show why it works in general. you seem to be begging the question here

5. Originally Posted by princess08
1. Prove that
a < b and c < 0 implies that b*c < a*c.
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