103k views
4 votes
Problem:

Prove the following:
(a) Suppose that a; b > 0. If a < b; then a2 < b2: (b) Suppose that a; b > 0. If a2 < b2; then a < b:

1 Answer

2 votes

Answer:

a)

We know that:

a, b > 0

a < b

With this, we want to prove that a^2 < b^2

Well, we start with:

a < b

If we multiply both sides by a, we get:

a*a < b*a

a^2 < b*a

now let's go back to the initial inequality.

a < b

if we now multiply both sides by b, we get:

a*b < b*b

a*b < b^2

Then we have the two inequalities:

a^2 < b*a

a*b < b^2

a*b = b*a

Then we can rewrite this as:

a^2 < b*a < b^2

This means that:

a^2 < b^2

b) Now we know that a.b > 0, and a^2 < b^2

With this, we want to prove that a < b

So let's start with:

a^2 < b^2

only with this, we can know that a*b will be between these two numbers.

Then:

a^2 < a*b < b^2

Now just divide all the sides by a or b.

if we divide all of them by a, we get:

a^2/a < a*b/a < b^2/a

a < b < b^2/a

In the first part, we have a < b, this is what we wanted to get.

Another way can be:

a^2 < b^2

divide both sides by a^2

1 < b^2/a^2

Let's apply the square root in both sides:

√1 < √( b^2/a^2)

1 < b/a

Now we multiply both sides by a:

a < b

User Chris Thornhill
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories