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Verify the rule that for two real numbers X and Y then

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Verify the rule that for two real numbers X and Y then |X+Y|≤|X|+|Y|



    2. Relevant equations
    3. The attempt at a solution

    1. When all the variables are positive:

    |X+Y|=(X+Y) because (X+Y)>0

    |X|=X because X>0

    |Y|=Y because Y>0

    So we got

    (X+Y)=X+Y

    2. When all the variables are negative:

    |-X+(-Y)|= -(-[X+Y])= (X+Y) because -(X+Y)<0

    |-X|= -(-X)=X because -X<0

    |-Y|= -(-Y)=Y because -Y<0

    Now we got

    (X+Y)=X+Y

    3. When the two variables have opposite signs:

    |X+(-Y)|= (X-Y) because (X-Y)>0

    |X|=X because X>0

    |-Y|= -(-Y)=Y because Y<0

    Finally we got:

    (X-Y)< X+Y

    ==================================================

    So that solution was shot down as a bit shaky. How about the one below then?

    |X+Y|≤|X|+|Y|

    |X+Y|-|X|-|Y|≤0

    If we remove the variables from the modulus we get

    X+Y-X-Y≤0

    Thanks.
     
  2. jcsd
  3. Nov 10, 2011 #2
    Not sure if this works

    |X+Y|≤|X|+|Y|

    |X+Y|-|X|-|Y|≤0

    0≤ X+Y-X-Y≤ 0

    0≤ 0≤ 0
     
  4. Nov 10, 2011 #3

    Mark44

    Staff: Mentor

    Here's a better way to say this.
    If x >= 0, and y >= 0,

    |x + y| = x + y = |x| + |y|
    You need to show, when x < 0 and y < 0, that |x + y| <= |x| + |y|. It looks like you showed (proved) that (x + y) = x + y. This is true, but not what you're supposed to show.

    Start with |x + y| and show that this equals |x| + |y|.

    Similar comments as above in your case 2. It is in these two cases (3 and 4) that you'll end up with actual inequalities.
    The thing is, you're not being asked about |x + (-y)| - you need to show that |x + y| < |x| + |y|.
     
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