Absolute Values and Intervals

MATH 133 – Calculus with Fundamentals, section 1, Prof. Little

Absolute Value

June 10, 2015 In everyday terms, the absolute value leaves nonnegative numbers unchanged, but “flips” a negative number to the corresponding positive number. The “flipping” can be done by multiplying by −1, so as a formula, if x is a real number, then

|x| = \begin{cases}x & if x ≥ 0 \\ −1 · x = −x & if x < 0. \end{cases}

For instance, since 15.9 and 0 are both ≥ 0,

|15.9| = 15.9 and |0| = 0

but since −8.3 < 0,

| − 8.3| = −(−8.3) = +8.3.

In everyday terms, the absolute value leaves nonnegative numbers unchanged, but “flips” a negative number to the corresponding positive number. The “flipping” can be done by multiplying by −1, so as a formula, if x is a real number, then

|x| = \begin{cases}x & if x ≥ 0 \\ −1 · x = −x & if x < 0. \end{cases}

For instance, since 15.9 and 0 are both ≥ 0,

|15.9| = 15.9 and |0| = 0

but since −8.3 < 0,

| − 8.3| = −(−8.3) = +8.3.

In everyday terms, the absolute value leaves nonnegative numbers unchanged, but “flips” a negative number to the corresponding positive number. The “flipping” can be done by multiplying by −1, so as a formula, if x is a real number, then

|x| = \begin{cases}x & if x ≥ 0 \\ −1 · x = −x & if x < 0. \end{cases}

For instance, since 15.9 and 0 are both ≥ 0,

|15.9| = 15.9 and |0| = 0

but since −8.3 < 0,

| − 8.3| = −(−8.3) = +8.3.

Geometric meaning

We know that the collection of all real numbers can be seen as the set of points on the number line.

[figure: 0]

If x ≥ 0, then |x| is the distance from 0 to x along the line:

[figure: 0; x; |x|]

If x < 0, |x| is still the positive distance along the line:

[figure: 0; x; |x|]

Absolute Values and Intervals

Geometric meaning

We know that the collection of all real numbers can be seen as the set of points on the number line.

[figure: 0]

If x ≥ 0, then |x| is the distance from 0 to x along the line:

[figure: 0; x; |x|]

If x < 0, |x| is still the positive distance along the line:

[figure: 0; x; |x|]

Absolute Values and Intervals

Similarly, ...

If a and b are any two real numbers, then |a − b| is the (positive) distance along the number line from a to b (or what is the same, from b to a, if the numbers satisfy b < a as in the following picture):

[figure: a; b; |a − b|]

For instance, |(−3) − 2| = | − 5| = 5 gives the distance from b = −3 to a = 2 along the number line.

Intervals

We will use this notation for intervals on the real number line: The closed interval [a, b] is the set of all real x with a ≤ x and x ≤ b. It is more common to write a ≤ x ≤ b here, but you should always understand that this is the same as the two separate inequalities a ≤ x and x ≤ b given first. The open interval (a, b) is the set of all real x with a < x and x < b, or a < x < b (not including the endpoints x = a, b). The half-closed interval [a, b) includes the endpoint a (and all x strictly between a and b), but not b. Similarly, the half-closed interval (a, b] includes the endpoint b, but not a.

An example

Let’s ask: What set of real numbers x is defined by the condition |x − 2| < 3? In geometric terms, the distance d from x to 2 along the number line is strictly less than 3. So x must be less than 5 and greater than −1:

[figure: 2; 5; −1; d = 3; d = 3]

That is, |x − 2| < 3 says the same thing as −1 < x < 5, so our set is the open interval (−1, 5).

Absolute Values and Intervals

Let’s ask: What set of real numbers x is defined by the condition |x − 2| < 3? In geometric terms, the distance d from x to 2 along the number line is strictly less than 3. So x must be less than 5 and greater than −1:

[figure: 2; 5; −1; d = 3; d = 3]

That is, |x − 2| < 3 says the same thing as −1 < x < 5, so our set is the open interval (−1, 5).

Absolute Values and Intervals

Let’s ask: What set of real numbers x is defined by the condition |x − 2| < 3? In geometric terms, the distance d from x to 2 along the number line is strictly less than 3. So x must be less than 5 and greater than −1:

[figure: 2; 5; −1; d = 3; d = 3]

That is, |x − 2| < 3 says the same thing as −1 < x < 5, so our set is the open interval (−1, 5).

Absolute Values and Intervals

Another way to see this with algebra

For any real number a, the statement |a| < r says the same thing as −r < a < r (think of the distance from a to 0). So

|x − 2| < 3 is the same as −3 < x − 2 < 3,
which is the same as (−3) + 2 < x < 3 + 2,
which is the same as −1 < x < 5.

So as before, the set of all such x is the interval (−1, 5).

Absolute Values and Intervals

Any interval can be described with absolute values

In fact, any open interval (a, b) can be described in the form we saw in the last example. To illustrate this, let’s consider the example of (4, 7). We want to write “the set of all x with 4 < x < 7” in the form “the set of all x with |x − c| < r for some c and r.” Thinking about what happened before, we see that c should be exactly half-way between 4 and 7. Thus, c is the midpoint of the interval: c = \frac{4+7}{2} = \frac{11}{2} = 5.5. Then, r is the distance from the midpoint to either endpoint: |7 − 5.5| = 1.5 and |4 − 5.5| = 1.5. Conclusion: The interval (4, 7) is the set of x with |x − 5.5| < 1.5.

In fact, any open interval (a, b) can be described in the form we saw in the last example. To illustrate this, let’s consider the example of (4, 7). We want to write “the set of all x with 4 < x < 7” in the form “the set of all x with |x − c| < r for some c and r.” Thinking about what happened before, we see that c should be exactly half-way between 4 and 7. Thus, c is the midpoint of the interval: c = \frac{4+7}{2} = \frac{11}{2} = 5.5. Then, r is the distance from the midpoint to either endpoint: |7 − 5.5| = 1.5 and |4 − 5.5| = 1.5. Conclusion: The interval (4, 7) is the set of x with |x − 5.5| < 1.5.

In fact, any open interval (a, b) can be described in the form we saw in the last example. To illustrate this, let’s consider the example of (4, 7). We want to write “the set of all x with 4 < x < 7” in the form “the set of all x with |x − c| < r for some c and r.” Thinking about what happened before, we see that c should be exactly half-way between 4 and 7. Thus, c is the midpoint of the interval: c = \frac{4+7}{2} = \frac{11}{2} = 5.5. Then, r is the distance from the midpoint to either endpoint: |7 − 5.5| = 1.5 and |4 − 5.5| = 1.5. Conclusion: The interval (4, 7) is the set of x with |x − 5.5| < 1.5.

In fact, any open interval (a, b) can be described in the form we saw in the last example. To illustrate this, let’s consider the example of (4, 7). We want to write “the set of all x with 4 < x < 7” in the form “the set of all x with |x − c| < r for some c and r.” Thinking about what happened before, we see that c should be exactly half-way between 4 and 7. Thus, c is the midpoint of the interval: c = \frac{4+7}{2} = \frac{11}{2} = 5.5. Then, r is the distance from the midpoint to either endpoint: |7 − 5.5| = 1.5 and |4 − 5.5| = 1.5. Conclusion: The interval (4, 7) is the set of x with |x − 5.5| < 1.5.

In fact, any open interval (a, b) can be described in the form we saw in the last example. To illustrate this, let’s consider the example of (4, 7). We want to write “the set of all x with 4 < x < 7” in the form “the set of all x with |x − c| < r for some c and r.” Thinking about what happened before, we see that c should be exactly half-way between 4 and 7. Thus, c is the midpoint of the interval: c = \frac{4+7}{2} = \frac{11}{2} = 5.5. Then, r is the distance from the midpoint to either endpoint: |7 − 5.5| = 1.5 and |4 − 5.5| = 1.5. Conclusion: The interval (4, 7) is the set of x with |x − 5.5| < 1.5.

In fact, any open interval (a, b) can be described in the form we saw in the last example. To illustrate this, let’s consider the example of (4, 7). We want to write “the set of all x with 4 < x < 7” in the form “the set of all x with |x − c| < r for some c and r.” Thinking about what happened before, we see that c should be exactly half-way between 4 and 7. Thus, c is the midpoint of the interval: c = \frac{4+7}{2} = \frac{11}{2} = 5.5. Then, r is the distance from the midpoint to either endpoint: |7 − 5.5| = 1.5 and |4 − 5.5| = 1.5. Conclusion: The interval (4, 7) is the set of x with |x − 5.5| < 1.5.

In fact, any open interval (a, b) can be described in the form we saw in the last example. To illustrate this, let’s consider the example of (4, 7). We want to write “the set of all x with 4 < x < 7” in the form “the set of all x with |x − c| < r for some c and r.” Thinking about what happened before, we see that c should be exactly half-way between 4 and 7. Thus, c is the midpoint of the interval: c = \frac{4+7}{2} = \frac{11}{2} = 5.5. Then, r is the distance from the midpoint to either endpoint: |7 − 5.5| = 1.5 and |4 − 5.5| = 1.5. Conclusion: The interval (4, 7) is the set of x with |x − 5.5| < 1.5.

The general pattern

If a < b, then the open interval (a, b): has midpoint c = \frac{a+b}{2}, and the distance from c to either endpoint is

|b − \frac{a + b}{2} | = \frac{b − a}{2} = |a − \frac{a + b}{2} | .

So the interval is the set of all x with

|x − \frac{a + b}{2} | < \frac{b − a}{2}.

No need to memorize this formula, but understand the process of how we got it!

Absolute Values and Intervals