2 part inequality problem, help please?

Question

An understanding would be great on how to solve a problem like this.

It has two parts (a, and b) to it.

a.) Suppose A and B are positive and A < B. Explain how you know that A^2 < B^2.

b.) Suppose you know that for two numbers A and B, A^2 < B^2. Does it follow that A < B? Discuss the different cases depending on which of the two numbers A and B is positive.


Thanks in advance, it means a bunch. (I am NOT just scrounging for answers, I would like an understanding, although answers would be extremely helpful to see how they are gotten. Thanks.)

Answer 1

If A is maller than B, A square will be smaller than B square. For instance:

Let's assume A = 3 and B = 4

3 < 4

3^2 = 3 *3 = 9

4^2 = 4*4 = 16

9 < 16

See whatever number you pick the result will be same. Because you are multiplying the numbers with themselves A will always b smaller than B.

Same as the second case. Let's pick two different numbers:

A^2 < B^2

10^2 < 100^2

See that gives you

10 * 10 = 100 and

100 * 100 = 10000 as a result

100 < 10000.

Hope this helps. Good Luck!

PS: Lonetree ^ means square, whcih you should know in order to answer this question.

Answer 2

A) The function x->f(x)=x^2 is increasing when x>0, which means that for 0
B) You have to note that the same function 'f' is also decreasing when x<0, which means that for Af(B)>f(0). So if you know that A^2 - If A and B positive, then A - If A and B negative, then AB)
- If A and B have different sign, all bets are off, since it could be either way, AB - for instance A=-1B=-3, yet you have 1<9 in both cases.

Answer 3

a. If we know that both A and B are positive, and that B>A, we know that 0 Therefore there is a positive number, x, such that A + x = B.
Given that A A AA < A(A+x) < (A+x)(A+x) since A+x is greater than A
A^2 < B^2 by substitution since A+x=B

b. We know that it does not follow that A A^2=1, B^2=4
So B^2 is greater than A^2, but B is not greater than A.

I will leave the various cases to be shown for you to discover.

Answer 4

1.since in inequalities operation with positive nos are exactly like in equations if A and B are positive and A =>A^2 2.if A^2 is consider A is 3and B is -5
(3)^2=9 and (-5)^2=25
so A^2 but A>B

Answer 5

to understand the inequality, you need to know the curve

y = x ^2

this is a u shaped curve touching the origin. On the positive side ie to the right of the origin, y is higher if the number is larger on the x-axis.

on the negative side ie to the left of the origin, y is higher if the number is smaller (more negative) on the x-axis.

Hence the converse is not true in turning from part b.

hope this helps.. you may want to search internet for sketch curves y = x ^ 2

Answer 6

a)
If A>0 & B>0 & A
E!k>0; kA=B =>
A A
A^2 Here E! mean: there exist exact one

b)

(A,B) belong to (RxR)
A^2 A
...lAl A
Case 1:
if A≥0 & B≥0 =>
(lAl A ==================

Case 2
if A≥0 & B≤0 =>
(lAl A ==================

Case 3
if A≤0 & B≤0 =>
(lAl A
(A>B => A ================
I have multiplyed the first term with -1 in order to get rid of (l l)=absolute value

Case 4
if A≤0 & B≥ 0=>
(lAl A =================
Conclusion:
It's not possible to deduce that :
A^2 A

Answer 7

your fortunate i admire doing those... a million. -9<4x+3<11 -9<4x-8<0 -a million<4x<0 -.253x-a million>-10 14>3x+9>0.5>3x>0 a million.666(r)>x>0 3. 2(x-a million)<-4 2x-2<-4 2x<-2 x<-a million thats it get somone else to do the rest...oh and that i could no longer be appropriate :P

Answer 8

A.) You are doing ^2 (whatever that is) to both of the numbers, so they cancel eachother out. So that just leaves you with A and B.

B.) Same answers as A.