Absolute value function, modulus function

Question

|x|+|y| = |x+y| then xy>0

|x|-|y|  =|x-y| then xy < 0

 

Can you please explain the above absolute value function properties and give some examples on the same 

Answer 1

|x| is always positive, no matter what, whether we started with +x or -x.

|y| is always positive, no matter what, whether we started with +y or -y.

 

Adding or subtracting these without the absolute values, you get one of four possible outcomes:

(+x) + (+y)       or    (+x) + (-y)        or  (-x) + (+y)          or      (-x) + (-y)

 

With both having the same sign  (xy>0) this first statement is always true, it is even true for x>y in the second statement.  But the second statement, |x| - |y| = |x - y| then xy<0 seems to be false.

 

Dealing with the addition first...   We will use x=3 and y=10 for our examples:

Both negative:     -3 + (-10) = -13.        

Both positive:  3 + 10 = 13.    

So as long as both numbers have the same sign (that is what this:   xy>0    means) the answer will ±13. Once you put this inside the absolute value, both results are positive 13 and |x| + |y| = |x+y|

 

Meaning   |-3| + |-10| = 3 + 10 = 13  (individual abs values)  and  

               | (-3) + (-10)| = |-13| = 13  (combined into one abs vaule)  (13 = 13 so works)

   |x| + |y| = | x + y |  if both numbers have the same sign  (xy>0)

 

Here it is with both numbers positive:

                        |3| + |10| = 3 + 10 = 13   and    | 3+10 | = |13| = 13  (13 = 13 so works)

 

 

 

On the other hand, if one is positive and the other negative , you are in essence subtracting.

 

For example  -3+10 = 7   or   3 + (-10) = -7

With different signs you get either +7 or -7 depending on the order you perform the subtraction.   If you put either of these results in an absolute value, you get the POSITIVE 7.

 

All four cases with subtraction when x<y do not work:

 

Both positive:  |3| - |10| = 3 - 10 = -7     and together inside    | 3 - 10 | = | -7 | = 7   (-7≠7 so not work!)

Both negative:  |-3| - |-10| = 3 - 10 = -7    and together inside    |-3 - (-10) | = | 7| = 7    (-7≠7 so not work!)

 

Differing signs  (xy<0):

The "3" neg and "10" pos:     |-3| - |10| = 3 - 10 = -7    and together inside  |-3 - 10 | = | -13 | = 13   (-7≠13 so not work!)

The "10" neg and "3" pos:     | 3| - | -10 | = 3 - 10 = -7    and together inside  |3 - (-10)| = |3 + 10 | = 13    (-7≠13 so not work!)

 

The second one, |x| - |y| = |x - y| when xy<0  doesn't work in any of the four cases while x<y. 

(notice:  |x| - |y| could give a negative result (when |x| < |y|), while | x - y | will always be positive)

 

 

 

But, if we require that  x > y, so now x=10 and y=3, then we have this:

 

Both pos:  |10| - |3| = 10 - 3 = 7   and   | 10 - 3 | = | 7 | = 7    (7 = 7 so works)

Both neg:  |-10| - |-3| = 10 - 3 = 7     and   |-10 - (-3) | = | -10 + 3 | = | -7 | = 7      (7 = 7 so works)

The "10" neg and "3" pos:   |-10| - |3| = 10 - 3 = 7       and     |-10 - (+3) | = |-13 | =  13    (7≠13 so NOT WORK!)

The "10" pos and "3" neg:  |10| - | -=3 | = 10 - 3 = 7    and    |10 - (-3) | = |10 + 3 | = |13| = 13    (7≠13 so NOT WORK!)

 

So the second statement of the original problem should probably say xy>0  AND  x>y.

 

I found this online - the true statement is:

 

|x-y|>|x|-|y| for xy<0.

 

This INEQUALITY is true, the the EQUATION you have on the second line is not true.

Answer 2

If |x|+|y| = |x+y| then xy>0. xy > 0 means that both x & y have the same signs. We assume that neither x nor y is zero, because in this case the proposition is obvious.

 

Explanation:If x and y have the same signs, then, in the expression |x+y| = (x+y), if both are positive, or = -(x+y), if both are negative.

 

In the latter case, |x|+|y| = -x + (-y) = -(x+y) , that is, the absolute value of x is -x, the absolute value of y is -y, and this turns out to be equal to the -(x+y) which, in this case, is |x+y| since |-(x+y)| = |x+y|.

 

The second proposition, where xy < 0, deals with differing-signed x and y. Furthermore, the absolute value of x must be greater than the absolute value of y, so that the quantity on the left (difference of two absolute values) is positive; this additional provision was omitted from what you presented.

 

You can work various examples to convince yourself of the correctness of these propositions, in specific illustrative cases.