4  Inequalities, Logical expressions

In this section we use the following package:

using CalculusWithJulia    # loads the `SpecialFunctions` package

4.1 Boolean values

In mathematics it is common to test if an expression is true or false. For example, is the point \((1,2)\) inside the disc \(x^2 + y^2 \leq 1\)? We would check this by substituting \(1\) for \(x\) and \(2\) for \(y\), evaluating both sides of the inequality and then assessing if the relationship is true or false. In this case, we end up with a comparison of \(5 \leq 1\), which we of course know is false.

Julia provides numeric comparisons that allow this notation to be exactly mirrored:

x, y = 1, 2
x^2 + y^2 <= 1

false

The response is false, as expected. Julia provides Boolean values true and false for such questions. The same process is followed as was described mathematically.

The set of numeric comparisons is nearly the same as the mathematical counterparts: <, <=, ==, >=, >. The syntax for less than or equal can also be represented with the Unicode ≤ (generated by \le[tab]). Similarly, for greater than or equal, there is \ge[tab].

Warning

The use of == is necessary, as = is used for assignment and mutation.

The ! operator takes a boolean value and negates it. It uses prefix notation:

!true

false

For convenience, a != b can be used in place of !(a == b).

4.2 Algebra of inequalities

To illustrate, let’s see that the algebra of expressions works as expected.

For example, if \(a < b\) then for any \(c\) it is also true that \(a + c < b + c\).

We can’t “prove” this through examples, but we can investigate it by the choice of various values of \(a\), \(b\), and \(c\). For example:

a,b,c = 1,2,3
a < b, a + c < b + c

(true, true)

Or in reverse:

a,b,c = 3,2,1
a < b, a + c < b + c

(false, false)

Trying other choices will show that the two answers are either both false or both true.

Warning

Well, almost… When Inf or NaN are involved, this may not hold, for example 1 + Inf < 2 + Inf is actually false. As would be 1 + (typemax(1)-1) < 2 + (typemax(1)-1).

So adding or subtracting most any finite value from an inequality will preserve the inequality, just as it does for equations.

What about addition and multiplication?

Consider the case \(a < b\) and \(c > 0\). Then \(ca < cb\). Here we investigate using \(3\) random values (which will be positive):

a,b,c = rand(3)   # 3 random numbers in [0,1)
a < b, c*a < c*b

(true, true)

Whenever these two commands are run, the two logical values should be identical, even though the specific values of a, b, and c will vary.

The restriction that \(c > 0\) is needed. For example, if \(c = -1\), then we have \(a < b\) if and only if \(-a > -b\). That is the inequality is “flipped.”

a,b = rand(2)
a < b, -a > -b

(false, false)

Again, whenever this is run, the two logical values should be the same. The values \(a\) and \(-a\) are the same distance from \(0\), but on opposite sides. Hence if \(0 < a < b\), then \(b\) is farther from \(0\) than \(a\), so \(-b\) will be farther from \(0\) than \(-a\), which in this case says \(-b < -a\), as expected.

Finally, we have the case of division. The relation of \(x\) and \(1/x\) (for \(x > 0\)) is that the farther \(x\) is from \(0\), the closer \(1/x\) is to \(0\). So large values of \(x\) make small values of \(1/x\). This leads to this fact for \(a,b > 0\): \(a < b\) if and only if \(1/a > 1/b\).

We can check with random values again:

a,b = rand(2)
a < b, 1/a > 1/b

(true, true)

In summary we investigated numerically that the following hold:

• a < b if and only if a + c < b + c for all finite a, b, and c.
• a < b if and only if c*a < c*b for all finite a and b, and finite, positive c.
• a < b if and only if -a > -b for all finite a and b.
• a < b if and only if 1/a > 1/b for all finite, positive a and b.

4.2.1 Examples

We now show some inequalities highlighted on this Wikipedia page.

Numerically investigate the fact \(e^x \geq 1 + x\) by showing it is true for three different values of \(x\). We pick \(x=-1\), \(0\), and \(1\):

x = -1; exp(x) >= 1 + x
x =  0; exp(x) >= 1 + x
x =  1; exp(x) >= 1 + x

true

Now, let’s investigate that for any distinct real numbers, \(a\) and \(b\), that

\[ \frac{e^b - e^a}{b - a} > e^{(a+b)/2} \]

For this, we use rand(2) to generate two random numbers in \([0,1)\):

a, b = rand(2)
(exp(b) - exp(a)) / (b-a) > exp((a+b)/2)

true

This should evaluate to true for any random choice of a and b returned by rand(2).

Finally, let’s investigate the fact that the harmonic mean, \(2/(1/a + 1/b)\) is less than or equal to the geometric mean, \(\sqrt{ab}\), which is less than or equal to the quadratic mean, \(\sqrt{a^2 + b^2}/\sqrt{2}\), using two randomly chosen values:

let
    a, b = rand(2)
    h = 2 / (1/a + 1/b)
    g = (a * b) ^ (1 / 2)
    q = sqrt((a^2 + b^2) / 2)
    h <= g, g <= q
end

(true, true)

4.3 Chaining, combining expressions: absolute values

The absolute value notation can be defined through cases:

\[ \lvert x\rvert = \begin{cases} x & x \geq 0\\ -x & \text{otherwise}. \end{cases} \]

The interpretation of \(\lvert x\rvert\), as the distance on the number line of \(x\) from \(0\), means that many relationships are naturally expressed in terms of absolute values. For example, a simple shift: \(\lvert x -c\rvert\) is related to the distance \(x\) is from the number \(c\). As common as they are, the concept can still be confusing when inequalities are involved.

For example, the expression \(\lvert x - 5\rvert < 7\) has solutions which are all values of \(x\) within \(7\) units of \(5\). This would be the values \(-2< x < 12\). If this isn’t immediately intuited, then formally \(\lvert x - 5\rvert <7\) is a compact representation of a chain of inequalities: \(-7 < x-5 < 7\). (Which is really two combined inequalities: \(-7 < x-5\) and \(x-5 < 7\).) We can “add” \(5\) to each side to get \(-2 < x < 12\), using the fact that adding by a finite number does not change the inequality sign.

Julia’s precedence for logical expressions, allows such statements to mirror the mathematical notation:

x = 18
abs(x - 5) < 7

false

This is to be expected, but we could also have written:

-7 < x - 5 < 7

false

Read aloud this would be “minus \(7\) is less than \(x\) minus \(5\) and \(x\) minus \(5\) is less than \(7\)”.

The “and” equations can be combined as above with a natural notation. However, an equation like \(\lvert x - 5\rvert > 7\) would emphasize an or and be “\(x\) minus \(5\) less than minus \(7\) or \(x\) minus \(5\) greater than \(7\)”. Expressing this requires some new notation.

The boolean shortcut operators && and || implement “and” and “or.” (There are also bitwise boolean operators & and |, but we only describe the former.)

Thus we could write \(-7 < x-5 < 7\) as

(-7 < x - 5) && (x - 5 < 7)

false

and could write \(\lvert x-5\rvert > 7\) as

(x - 5 < -7) || (x - 5 > 7)

true

(The first expression is false for \(x=18\) and the second expression true, so the “or”ed result is true and the “and” result is false.)

Example

One of DeMorgan’s Laws states that “not (A and B)” is the same as “(not A) or (not B)”. This is a kind of distributive law for “not”, but note how the “and” changes to “or”. We can verify this law systematically. For example, the following shows it true for \(1\) of the \(4\) possible cases for the pair A, B to take:

A,B = true, false  ## also true, true; false, true; and false, false
!(A && B) == !A || !B

true

4.4 Precedence

The question of when parentheses are needed and when they are not is answered by the precedence rules implemented. Earlier, we wrote

(x - 5 < -7) || (x - 5 > 7)

true

To represent \(\lvert x-5\rvert > 7\). Were the parentheses necessary? Let’s just check.

x - 5 < -7 || x - 5 > 7

true

So no, they were not in this case.

An operator (such as <, >, || above) has an associated associativity and precedence. The associativity is whether an expression like a - b - c is (a-b) - c or a - (b-c). The former being left associative, the latter right. Of issue here is precedence, as in with two or more different operations, which happens first, second, \(\dots\).

The table in the manual on operator precedence and associativity shows that for these operations “control flow” (the && above) is lower than “comparisons” (the <, >), which are lower than “Addition” (the - above). So the expression without parentheses would be equivalent to:

((x-5) < -7) || ((x-5) > 7)

true

(This is different than the precedence of the bitwise boolean operators, which have & with “Multiplication” and | with “Addition”, so x-5 < 7 | x - 5 > 7 would need parentheses.)

A thorough understanding of the precedence rules can help eliminate unnecessary parentheses, but in most cases it is easier just to put them in.

4.5 Arithmetic with

For convenience, basic arithmetic can be performed with Boolean values, false becomes \(0\) and true \(1\). For example, both these expressions make sense:

true + true + false, false * 1000

(2, 0)

The first example shows a common means used to count the number of true values in a collection of Boolean values - just add them.

This can be cleverly exploited. For example, the following expression returns x when it is positive and \(0\) otherwise:

(x > 0) * x

18

There is a built in function, max that can be used for this: max(0, x).

This expression returns x if it is between \(-10\) and \(10\) and otherwise \(-10\) or \(10\) depending on whether \(x\) is negative or positive.

(x < -10)*(-10) + (x >= -10)*(x < 10) * x + (x>=10)*10

10

The clamp(x, a, b) performs this task more generally, and is used as in clamp(x, -10, 10).

4.6 Questions

Question

Is e^pi or pi^e greater?

Select an item

e^pi is greater than pi^e

e^pi is less than pi^e

e^pi is equal to pi^e

Question

Is \(\sin(1000)\) positive?

Select an item

Yes

No

Question

Suppose you know \(0 < a < b\). What can you say about the relationship between \(-1/a\) and \(-1/b\)?

Select an item

\(-1/a > -1/b\)

\(-1/a < -1/b\)

\(-1/a \geq -1/b\)

Question

Suppose you know \(a < 0 < b\), is it true that \(1/a > 1/b\)?

Select an item

It can sometimes be true, though not always.

It is never true, as \(1/a\) is negative and \(1/b\) is positive

Yes, it is always true.

Question

The airyai function is a special function named after a British Astronomer who realized the function’s value in his studies of the rainbow. The SpecialFunctions package must be loaded to include this function, which is done with the accompanying package CalculusWithJulia.

airyai(0)

0.3550280538878172

It is known that this function is always positive for \(x > 0\), though not so for negative values of \(x\). Which of these indicates the first negative value : airyai(-1) <0, airyai(-2) < 0, …, or airyai(-5) < 0?

Select an item

airyai(-1) < 0

airyai(-2) < 0

airyai(-3) < 0

airyai(-4) < 0

airyai(-5) < 0

Question

By trying three different values of \(x > 0\) which of these could possibly be always true:

Select an item

x^x == (1/e)^(1/e)

x^x <= (1/e)^(1/e)

x^x >= (1/e)^(1/e)

Question

Student logic says \((x+y)^p = x^p + y^p\). Of course, this isn’t correct for all \(p\) and \(x\). By trying a few points, which is true when \(x,y > 0\) and \(0 < p < 1\):

Select an item

(x+y)^p == x^p + y^p

(x+y)^p < x^p + y^p

(x+y)^p > x^p + y^p

Question

According to Wikipedia, one of the following inequalities is always true for \(a, b > 0\) (as proved by I. Ilani in JSTOR, AMM, Vol.97, No.1, 1990). Which one?

Select an item

a^b + b^a <= 1

a^a + b^b >= a^b + b^a

a^a + b^b <= a^b + b^a

Question

Is \(3\) in the set \(\lvert x - 2\rvert < 1/2\)?

Select an item

Yes

No

Question

Which of the following is equivalent to \(\lvert x - a\rvert > b\):

Select an item

\(x - a < -b \text{ or } x - a > b\)

\(-b < x-a \text{ and } x - a < b\)

\(-b < x - a < b\)

Question

If \(\lvert x - \pi\rvert < 1/10\) is \(\lvert \sin(x) - \sin(\pi)\rvert < 1/10\)?

Guess an answer based on a few runs of

x = pi + 1/10 * (2rand()-1)
abs(x - pi) < 1/10, abs(sin(x) - sin(pi)) < 1/10

Select an item

true

false

Question

Does 12 satisfy \(\lvert x - 3\rvert + \lvert x-9\rvert > 12\)?

Select an item

Yes

No

Question

Which of these will show DeMorgan’s law holds when both values are false:

Select an item

!(false && false) == (false || false)

!(false && false) == (!false && !false)

!(false && false) == (!false || !false)

Question

For floating point numbers there are two special values Inf and NaN. For which of these is the answer always false:

Select an item

NaN < 3.0 and 3.0 <= NaN

Inf < 3.0 and 3.0 <= Inf

Question

The IEEE 754 standard is about floating point numbers, for which there are the special values Inf, -Inf, NaN, and, surprisingly, -0.0 (as a floating point number and not -0, an integer). Here are 4 facts that seem reasonable:

• Positive zero is equal but not greater than negative zero.
• Inf is equal to itself and greater than everything else except NaN.
• -Inf is equal to itself and less then everything else except NaN.
• NaN is not equal to, not less than, and not greater than anything, including itself.

Do all four seem to be the case within Julia? Find your answer by trial and error.

Select an item

Yes

No

Question

The NaN value is meant to signal an error in computation. Julia has value to indicate some data is missing or unavailable. This is missing. For missing values we have these computations:

true && missing, true || missing

(missing, true)

We see the value of true || missing is true. Why?

Select an item

Since the second value is "missing", only the first is used. So false || missing would also be false

In the manual we can read that "In the expression a || b, the subexpression b is only evaluated if a evaluates to false." In this case a is true and so a is returned.

The value for true && missing is missing, not a boolean value. What happens?

Select an item

In the manual we can read that "In the expression a && b, the subexpression b is only evaluated if a evaluates to true." In this case, a is true so b is evaluated and returned. As b is just missing that is the return value.

Since the second value is "missing" all such answers would be missing.