This is a special case of the magnitude of a complex number.
Think of points and on a number line and assume that and
Comparing the two numbers, we can easily say simply because is positive and is negative. What if we do not care about the sign but only the distance of each number from zero? Then we say or where is the absolute value notation that gives the respective values of and without regard to their signs. Hence the following definition:
For any real number we define its absolute value as follows:
Then why absolute values? When do we consider only the deviation from zero? Suppose your house is located on an east-west road and your car parked in the driveway has been stolen. If the thief is close enough, you want to go out and catch him yourself. Otherwise, you will call the police. According to information gathered, the car thief is said to have driven away east for miles and, for whatever reason, have turned around and driven west for miles.
Taking east as positive and west negative, the thief is currently mile away, i.e. his location is mile west of your house. However, remember that the only factor that you take into consideration as to whether to go out yourself or call the police is the thief's distance from your house regardless of the direction. Then the calculation relevant to you right at this moment would be as opposed to This is, for example, where absolute value plays a role. Make sense?
Now, let us think about something more interesting. Could the stolen car's distance from your house ever be greater than the distance the car traveled to get there? We know from above that the distance of the car from your house is mile. Since the distance traveled by the car is the answer to the question is no.
What if the thief traveled farther east, instead of turning back and heading west, for miles? Would your answer still be no? Let us see. In this case, the distance of the car from your house in miles is whereas the distance traveled by the car in miles is
Since these two numbers are equal, the answer to the above question is still no. That is, your stolen car's distance from your house can never be greater than the distance the car traveled to get there. In general, for any real numbers and which is part of the following:
Properties of absolute value:
- if and only if or
- for any positive integer
Here are some examples of how to calculate absolute values using the above properties:
The quantities and are positive, so they remain unchanged when the absolute bars are dropped. However, because is negative, we have
Find the value of Express your answer as a fraction in its lowest term.
Now, it is time for a graphical interpretation of absolute value. Take a look at the following graph of
Do you see that it is drawn exactly by the definition of absolute value and the value of is always non-negative? Do you see that the dotted red line, which is for is flipped over against the -axis so that the negative values of become positive?
As regards to the car thief example, let us say that the car owner wants to go out and catch the thief himself if the bad guy is within miles from his house. Then in the first scenario where the thief ends up mile west of his house and thus the thief is mile away from his house, he will not call the police but go out himself. On the contrary, in the second scenario where the thief ends up miles east of his house and thus the thief is miles away from his house, he will call the police.
Now, what if the car owner's house was located at instead of in the first place and the car thief behaved in the same manner? How different will the graph look? The answer is right below:
Since the house is located at under this new circumstance, the thief in the first scenario does not end up but miles east of and thus his distance from the owner's house is mile. This would make sense to you if you noticed that in the above example can be considered as
In the second scenario, the thief does not end up but miles east of and thus his distance from the owner's house is miles. Hence the equation of the above graph is where the dotted red line, which is the graph of for is flipped over against the -axis at because the benchmark point is now instead of
Now that we know how to get the graph of which is -shaped, let us try to get a -shaped graph. To do that, we first translate the graph of by in the negative direction of the -axis, as shown below, and the equation of the translated graph is
Then we ask, "What will the graph of look like?" Since all the values of for in the graph of are negative, we flip over that part of the graph to obtain the following, just as we did above:
Do you really see a -shaped graph now? One question that naturally arises in your mind would be how to draw the graph of from scratch without referring to the thief example. We can use linear inequalities to accomplish this.
Observe from the equation that if then This can be rewritten in more detail as the following two cases: Can you confirm that corresponds to the above graph for and corresponds to the above graph for I think you already did.
Similarly, observe from the equation that if then which is equivalent to This can be rewritten in more detail as the following two cases: Again, can you confirm that corresponds to the above graph for and corresponds to the above graph for I am sure you already did.
Now, it's time for you to try some examples.
Which of the following is the graph of for
If then which implies If then which implies Therefore, the correct answer is
Which of the following conditions does NOT always satisfy
(a) is a non-negative real number.
(b) is a positive real number.
(c) is a positive integer.
(d) is a non-zero real number.
If then which implies holds for all non-negative real numbers. Hence, and always satisfy
If then which implies never holds for
Therefore, the correct answer is
Given that are non-zero real numbers , find all possible values of the expression .
Since for any and for any ,
if are all negative;
if exactly two of are negative;
if exactly one of is negative;
if are all positive.
Thus , the possible values of the given expression are and
Solve the equation
We need to discuss three cases:
any is a solution.
But since we assumed , there is no solution for the case .
In conclusion, the solutions to the equation are .
Find the minimum value of .
Case 1 : When ,
Case 2: When ,
Case 3: When ,
Case 4: When ,
Thus the global minimum of the given expression is 4.