# Inequality problems

1. Jun 27, 2011

### Curd

attached are the problems (actually i don't think i bothered with #96) i'm having trouble with.

attached is ONE of my attempts

and attached is the book's answers.

I have NO idea where to even begin with these.

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2. Jun 27, 2011

### Staff: Mentor

For 97, you have a note for 0 <= |a| that says this is not true. It most certainly is true.

For 101, where you have a question, the assumption is that x < -sqrt(a), so x + sqrt(a) < 0.
So x - sqrt(a) < -sqrt(a) - sqrt(a) = -2sqrt(a) < 0. This shows that x - sqrt(a) < 0.
Now, since x + sqrt(a) < 0, as well, the product (x - sqrt(a))(x + sqrt(a)) > 0, since both factors are negative.

3. Jun 27, 2011

### Curd

how do you get that for 97? a could be anything. and how can a number have an absolute value that is less than 0 and = to 0? the number must be the same distance from 0 on both sides of 0.

i have no idea what you've done with 101. it looks like you've arbitrarily tossed some equations together.

Last edited: Jun 27, 2011
4. Jun 27, 2011

### Staff: Mentor

You are misreading the inequality in two ways. The inequality 0 <= |a| says
1) |a| is greater than 0 (not less than, as you wrote).
OR (not "and", as you wrote).
2) |a| is equal to zero

5. Jun 27, 2011

### Curd

this is what i got for 101. i can't seem to get one of the inequality symbols to turn the right direction. what did i do wrong?

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• ###### prob 101.jpg
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6. Jun 27, 2011

### Curd

1) okay, so how can a number have an absolute value that is GREATER than 0 and = to 0? the number must be the same distance from 0 on both sides of 0.

2) how can an absolute be equal to 0 and less than OR greater than 0? the only thing i can think of is -0 and +0 which are both still 0. the fact that one number must be zero implies that the absolute value has values, one of which is 0 and one of which is greater than 0. this is impossible since both numbers of the absolute value must be the same distance from 0. as in -1 and +1 or -8 and +8.

3) i can show that |1| is 1 and -1 though... and that |8| is 8 and -8 ergo since |a| is a and -a then |a| is equal to or less than a since a is a and -a is less than a. but i'm guessing that's not really where they were going with that problem?

Last edited: Jun 27, 2011
7. Jun 27, 2011

### Staff: Mentor

"and" is the wrong word. I hoped that by writing it in red (OR) you would notice that. Apparently not.
It can't. A number can be:
less than zero
less than or equal to zero
equal to zero
greater than zero
greater than or equal to zero

A number can't be equal to zero AND less than zero OR greater than zero.
You are confused about the meaning of absolute value. The absolute value of a number is one value, not two. Your confusion might be that |1| = 1 while |-1| = 1 as well. For a given positive number, there are two real numbers whose absolute value is that number.
No you can't. |1| = 1. Period.
No. |8| = 8 and |-8| = 8.
That's wrong, too.

|a| = a if a happens to be positive, but |a| = -a, if a happens to be negative, but |a| represents only one value, not two as you are writing.

8. Jun 27, 2011

### Staff: Mentor

It would be easier if put your work here in text, rather than as a screen shot.

To comment on your work I have to use two screens, and switch back and forth between the two.

Where you have
(x - sqrt(a))(x + sqrt(a)) > 0,

you have two factors whose product is positive. For the sake of simpliciy, let's call them M and N.

If M*N > 0, then one of two things must be true:
1) M > 0 AND N > 0. (For example, (2)(6) = 12 > 0)
OR
2) M < 0 AND N < 0. (For example, (-2)(-6) = 12 > 0)

After this line -
(x - sqrt(a))(x + sqrt(a)) > 0

your work should include two cases, one where both factors are positive, and the other where both factors are negative.

9. Jun 27, 2011

### Curd

i haven't read your posts yet, but i just realized that |a| must be greater than or = to 0 because there can be no greater than or lesser than with |0|. the absolute value of 0 is just zero. so the absolute value, as in the value you get when you count the spaces from 0 on the number line, must always be more or equal to 0.

10. Jun 27, 2011

### Curd

then why is it that when we solve for absolute values we always solve for the negative and the positive?

11. Jun 27, 2011

### gb7nash

Let's say you want to solve for x, where |x| < 10. So in this case, -10 < x < 10. This makes sense since any value a between -10 and 10 will satisfy |a| < 10. e.g. |-9| = 9 < 10, |3| = 3 < 10, etc..

-----------

To answer your original question though, the absolute value of a number is defined as the numerical value, without regard to its sign.

12. Jun 27, 2011

### Staff: Mentor

This is where your confusion is. Any given real number has only one absolute value. After all, that number is a certain distance from zero. For any distance except zero, there are two numbers that distance away from zero.

The equation |x| = 5 has two solutions: x = 5 OR x = -5. Both of these numbers are 5 units away from zero.

13. Jun 28, 2011

### Curd

can we start this again with problem 97 and go in order through each problem? they are dependent on each other.

this is my new answer for number 97

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• ###### 97 again.jpg
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14. Jun 28, 2011

### Staff: Mentor

Again, it would be helpful if you posted your work directly on this page rather than a screen shot. As already mentioned, anyone helping needs to keep track of two screens. For another thing, to comment on your work, I have to type it, too, in addition to comments I make.

If you want help, don't make it more difficult for someone to give you help.

In the work you posted, there are several things that aren't true.

You wrote 0 <= |a|. This is true by definition. Your reason is incorrect, though.
"because a must be a positive value" - a can be any real number. |a| can be either 0 or positive, but it can't be negative.

"if absolute value of a is on the negative side ..." - |a| is never negative.

"if it is on the positive side the (sic) a is = to a" -- Any number is always equal to itself. What's the point of saying this?

15. Jul 5, 2011

### Curd

the problem is to prove a < or = to |a|

a can be a or -a but |a| is always going to be the positive of those two as it is the distance that is counted in the case of |a| and not the actual value. the point is to show that the a in |a| can be a or -a. so a is less than or equal to |a|. i'm not sure that's a proof but it is a given point of the definition i believe.

if this explanation is a correct solution to the problem then please show me number 98. if not then please critique it and the next problem too.

98) prove that |a+b| less than or equal to |a|+|b|

they give a hint to expand "|a+b|^2 = (a+b)^2" for some reason that i am not capable of understanding at the moment.

16. Jul 5, 2011

### eumyang

In (Elementary) Algebra you should have learned the "square of a binomial pattern" or "perfect square trinomial." If you do not recall this, remember that
(a + b)2 = (a + b)(a + b)
and you can multiply the two binomials using FOIL.

17. Jul 6, 2011

### Curd

um, i've known that for quite some time. that wasn't the issue.

the issue was, how am i to prove "that |a+b| less than or equal to |a|+|b" with that or without that.

18. Jul 6, 2011

### SammyS

Staff Emeritus
$$|x|=\left\{\begin{array}{cc}x,&\mbox{ if } x\geq 0\\-x, & \mbox{ if } x<0 \end{array}\right.$$

19. Jul 6, 2011

### eumyang

Well, your comment wasn't clear to me, then. You should have specified that you know about the square of a binomial pattern but didn't know how to use it.

Show us the expanded form of (a + b)2:
|a + b|2 = (a + b)2 = ????

...and then answer this: what can you say about a2 and |a|2? And after answering that, what can you do about the expanded trinomial?