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Absolute value in practice
Exercises on absolute value
Hello and welcome back to a step in practice– we are dealing here with absolute values. In this video, we are going to solve just exercises from 1 to 4, and we leave exercise 5 to the PDF. In exercise 1, we are asked to prove the triangle inequality– the absolute value of x plus y, less or equal than the sum of the absolute values. Now let us recall that the inequality, absolute value of x less or equal than a is equivalent to the fact that minus a is less than x, and at the same time, x is less than a.
This is the situation– if a is positive, the elements whose absolute value is less than or equal than a are just those between minus a and a. This is the case a positive. And if a is negative, strictly negative, there are no x satisfying the first inequality, and there are no x satisfying both of these inequalities. So if we put a equal to the sum of the absolute values, it’s enough to prove that x plus y lies between a and minus a. Now x is between its absolute value and the opposite of its absolute value. At the same time, y is between its absolute value and the opposite of the absolute value.
So if we take the sum, term by term, we get minus the sum of the two absolute values less or equal than x plus y, less or equal than the sum of the absolute values. And thus, this implies that the triangular inequality is true.
In exercise 2, we’ve got the function, which is the difference of two absolute values– precisely, the absolute value of x minus 1 and the absolute value of 4x plus 8. Now, x minus 1– while the absolute value of x minus 1 depends on the sign of x minus 1. And analogously, the absolute value of 4x plus 8 depends on the sign of 4x plus 8. Now x minus 1 is greater or equal than 0, if and only if x is greater than 1. Whereas 4x plus 8 is greater or equal than 0 if and only if 4x is greater or equal than 8. That is, x greater or equal than minus 2.
So we consider three cases, depending whether x is less than minus 2, between minus 2 and 1, or greater than 1. So if x is strictly less than minus 2, then 4x plus 8 is negative. And also x minus 1 is negative. And therefore f of x equals– well instead of the absolute values of x minus 1, we can write minus x minus 1, minus, minus, 4x plus 8. Which is minus x plus 1, plus 4x plus 8. That is 3x plus 9. If x belongs to the interval between minus 2 and 1, we take, for convenience, minus 2, and we do not take 1.
Then well, x is greater than minus 2, so 4x plus 8 is greater or equal than 0. But x minus 1 is negative. And thus, f of x equals minus x minus 1, minus 4x plus 8.
And this equals minus 5x plus 1, minus 8. So minus 7. Finally, if x is greater or equal than 1, then both terms are positive. 4x plus 8 is greater or equal than 0, x plus 1, x minus 1 is greater or equal than 0. And thus, we get f of x equal to x minus 1, minus 4x plus 8, which is exactly minus 3x, minus 9.
In exercise 3, we are asked to find the formula for the maximum between two real numbers– a and b.
Now notice that if we draw the segment between a and b, well, we can have b be on the right hand side or a on the right hand side. But in any case, there are some quantities that do not depend on the position of a with respect to b, like the midpoint within between a and b, which is always equal to a plus b over 2. And also the distance between a and b, which is always the absolute value of b minus a. We get the maximum between a and b by adding to the midpoint, half of the distance between a and b.
Therefore, the maximum between a and b is the midpoint plus the absolute value of b minus a over 2. And this is a formula involving just absolute values and sums and quotients. Now in exercise 4, we’ve three sets. Let us write them in a more compact way. The set A is the set of numbers from minus 3 to plus infinity, whereas the set B is the set of numbers that are strictly less than 6– so minus infinity, 6. And C is the set of numbers whose distance to 5 is less than or equal to 2. And so we get the closed interval 5 minus 2, 5 plus 2. That is, the interval 3, 7– with 3, and 7 on it.
Now it is convenient in order to find the intersection of the three sets to draw each set as a subset of the real line. So we draw 3 real lines, and on each of the real lines, we represent the sets A, B, and C. So the values that count here are minus 3, 3, 6, and 7. So we consider minus 3, 3, 6, and 7.
And in the first real line, we consider the set A, in the second set B, in the third the set C. And we draw one more line here with the intersection of A with B and C. So let us draw the set A is the set from minus 3 to plus infinity. And we take minus 3–
the set B goes from minus infinity to 6, but we do not take 6, so this is excluded.
And also the set C goes from 3 to 7, so we take a set from 3 to 7. And we take 7. So be careful. We do not take 6, but we take 7, 3 here, and minus 3 here. So the intersection is what is in common between the three sets. So we look in each column. Here, we do not take this column, we do not take this column, but we take this one. For sure we’ll take from 3 to 6, but be careful we do not take 6, and we take 3.
We do not take the column between 3, 6, and 7, and we don’t take the column between 7 plus infinity. So the intersection of the three sets is the set from 3– the interval from 3 to 6. We take 3, and we do not take 6. And this ends exercise 4. Exercise 5 is solved just in the PDF. So see you on the next step.
The following exercises are solved in this step.
We invite you to try to solve them before watching the video.
In any case, you will find below a PDF file with the solutions.
Exercise 1.
Prove the triangle inequality (|x+y|le |x|+|y|) for all (x,yinmathbb R)
Exercise 2.
Write (f(x) = |x-1| – |4x + 8|) as an expression where there are no absolute values.
Exercise 3.
Express the maximum (maxlbrace a,brbrace) between the real numbers (a) and (b) in a formula involving just sums, products and the absolute value of real numbers. (Hint: what is the distance from (maxlbrace a,brbrace) to the midpoint of the segment joining (a) to (b)?)
Exercise 4.
Let (A=lbrace xinmathbb R:, xge -3rbrace, B=lbrace xinmathbb R:, x<6rbrace,) ( C=lbrace xin mathbb R:, |x-5|le 2rbrace). Find the set (Acap Bcap C).
Exercise 5. [Solved only in the PDF file]
Knowing that (sqrt 3=1.732050808…) give an estimate of (|sqrt 3-1.73205|).
This article is from the online course:
Precalculus: the Mathematics of Numbers, Functions and Equations
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Precalculus: the Mathematics of Numbers, Functions and Equations
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