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Homework Help: Inequality problems

  1. Jun 27, 2011 #1
    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.

    Attached Files:

  2. jcsd
  3. Jun 27, 2011 #2


    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.
  4. Jun 27, 2011 #3
    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
  5. Jun 27, 2011 #4


    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
  6. Jun 27, 2011 #5
    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?

    Attached Files:

    Last edited: Jun 27, 2011
  7. Jun 27, 2011 #6
    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
  8. Jun 27, 2011 #7


    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.
  9. Jun 27, 2011 #8


    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)
    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.
  10. Jun 27, 2011 #9
    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.
  11. Jun 27, 2011 #10

    then why is it that when we solve for absolute values we always solve for the negative and the positive?
  12. Jun 27, 2011 #11


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    Homework Helper

    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.
  13. Jun 27, 2011 #12


    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.
  14. Jun 28, 2011 #13
    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

    Attached Files:

  15. Jun 28, 2011 #14


    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?
  16. Jul 5, 2011 #15
    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.
  17. Jul 5, 2011 #16


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    Homework Helper

    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.
  18. Jul 6, 2011 #17
    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.
  19. Jul 6, 2011 #18


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    It may help to start with the definition of absolute value.
    [tex]|x|=\left\{\begin{array}{cc}x,&\mbox{ if }
    x\geq 0\\-x, & \mbox{ if } x<0 \end{array}\right.[/tex]
  20. Jul 6, 2011 #19


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    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?
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