We have talked about quite a few concepts involving absolute value of x in our previous posts. But some absolute value questions involve two variables. Then do we need to consider the positive and negative values of both x and y? Certainly! But there are some properties of absolute value that could come in handy in such questions. Let’s take a look at them:

(I) For all real x and y, |x + y| <= |x| + |y|

(II) For all real x and y, |x – y| >= |x| – |y|

We don’t need to learn them of course and there is no need to look at how to prove them either. All we need to do is understand them – why do they hold, when is the equality sign applicable and when can they be useful. Let’s look at both the properties one by one.

**(I) For all real x and y, |x + y| <= |x| + |y|**

The result of both the left hand side and the right hand side will be positive or zero. On the right hand side, the absolute values of x and y will always get added irrespective of the signs of x and y. On the left hand side, the absolute values of x and y might get added or subtracted depending on whether they have the same sign or different signs. Hence the result of the left hand side might be smaller than or equal to that of the right hand side.

For which values of x and y will the equality hold and for which values will the inequality hold? Let’s think logically about it.

The absolute values of x and y get added on the right hand side. We want the absolute values of x and y to get added on the left hand side too for the equality to hold. This will happen when x and y have the same sign. So the equality should hold when they have the same signs.

For example, x = 4, y = 8:

|4 + 8| = |4| + |8| = 12

OR x = -3, y = -4:

|-3 -4| = |-3| + |-4| = 7

Also, when at least one of x and y is 0, the equality will hold.

For example, x = 0, y = 8:

|0 + 8| = |0| + |8| = 8

OR x = -3, y = 0:

|-3 + 0| = |-3| + |0| = 3

What happens when x and y have opposite signs? On the left hand side, the absolute values of x and y get subtracted hence the left hand side will be smaller than the right hand side (where they still get added). That is when the inequality holds i.e. |x + y| < |x| + |y|

For example, x = -4, y = 8:

|-4 + 8| < |-4| + |8|

4 < 12

OR x = 3, y = -4:

|3 -4| < |3| + |-4|

1 < 7

Let’s look at our second property now:

**(II) For all real x and y, |x – y| >= |x| – |y|**

Thinking on similar lines as above, we see that the right hand side of the inequality will always lead to subtraction of the absolute values of x and y whereas the left hand side could lead to addition or subtraction depending on the signs of x and y. The left hand side will always be positive whereas the right hand side could be negative too. So in any case, the left hand side will be either greater than or equal to the right hand side.

When will the equality hold?

When x and y have the same sign and x has greater (or equal) absolute value than y, both sides will yield a positive result which will be the difference between their absolute values

For example, x = 9, y = 2;

|9 – 2| = |9| – |2| = 7

OR x = -7, y = -3

|-7 – (-3)| = |-7| – |-3| = 4

Also when y is 0, the equality will hold.

For example, x = 8, y = 0:

|8 – 0| = |8| – |0| = 8

OR x = -3, y = 0:

|-3 – 0| = |-3| – |0| = 3

What happens when x and y have the same sign but absolute value of y is greater than that of x?

It is easy to see that in that case both sides have the same absolute value but the right hand side becomes negative.

For example, x = -4, y = -9

|x – y| = |-4 – (-9)| = 5

|x| – |y| = |-4| – |-9| = -5

So even though the absolute values will be the same since we will get the difference of the absolute values of x and y on both sides, the right hand side will be negative. If we were to take further absolute value of the right hand side, the two will become equal i.e. the right hand side will become |(|x| – |y|)| = |-5| = 5 in our example above. In that case, the equality will hold again.

Similarly, what happens when only x = 0? The right hand side becomes negative again so taking further absolute value will make both sides equal.

For example, x = 0, y = -5

|x – y| = |0 – (-5)| = 5

|x| – |y| = |0| – |5| = -5

Taking further absolute value, |(|x| – |y|)| = |-5| = 5

So when we take further absolute value of the right hand side, this property becomes similar to property 1 above: |x – y| = |(|x| – |y|)| when x and y have the same sign or at least one of x and y is 0.

Now let’s look at the inequality part of property 2.

Whenever x and y have opposite signs, |x – y| > |x| – |y|

On the left hand side, the absolute values will get added while on the right hand side, the absolute values will get subtracted. So the absolute value of the left hand side will always be greater than the absolute value of the right hand side. The left hand side will always be positive while the right hand side could be negative too. Hence even if we take the further absolute value of the right hand side, the inequality will hold: |x – y| > |(|x| – |y|)| when x and y have opposite signs

For example, x = -4, y = 8:

|-4 – 8| > |-4| – |8|

12 > -4

Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-4| = 4

Still, 12 > 4 i.e. |x – y| > |(|x| – |y|)|

OR x = 3, y = -4:

|3 –(-4)| > |3| – |-4|

7 > -1

Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-1| = 1

Still, 7 > 1 i.e. |x – y| > |(|x| – |y|)|

Note that the inequality of the original property 2 also holds when x and y have the same sign but absolute value of y is greater than the absolute value of x since the right hand side becomes negative. It also holds when x is 0 but y is not.

**To sum it all neatly,**

(I) For all real x and y, |x + y| <= |x| + |y|

|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x + y| < |x| + |y| when (1) x and y have opposite signs

(II) For all real x and y, |x – y| >= |x| – |y|

|x – y| = |x| – |y|when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0

|x – y| > |x| – |y| in all other cases

(III) For all real x and y, |x – y| >= |(|x| – |y|)|

|x – y| = |(|x| – |y|)| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x – y| > |(|x| – |y|)| when (1) x and y have opposite signs

Note that property (III) matches property (I).

There is another property we would like to discuss but let’s take it up next week along with some GMAT questions where we put these properties to use.

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*