Math and Multimedia

Why do we reverse/flip the inequality sign?

October 15, 2009 GB High School Algebra, High School Mathematics, Questions and Quandaries

You have probably remembered in Algebra that if we multiply an inequality by a negative number, then the inequality sign should be flipped or reversed. For example, if we want to find the solution of the inequality $-\frac{1}{2}x > 8$ , we multiply both sides by $-2$ and reverse the greater than sign giving us $x < -16$ . Now, why did the $>$ sign became $<$ ?

If we generalize the statements above, suppose we have two numbers, say, $a$ and $b$ such that $a > b$ , if we multiply them to a negative number $c$ , instead of having $ac > bc$ , the answer should be $ac < bc$ .

Before we proceed with our discussion, let us first remember 2 basic concepts we have learned in elementary mathematics:

The number line is arranged in such a way that the negative numbers are at the left hand side of $0$ and the positive numbers are at its right hand side such as shown in Figure 1.
If we have $2$ numbers $a$ and $b$ , then $a > b$ if $a$ is at the right of $b$ on the number line. For example, in Figure 1, $2 > -1$ since $2$ is at the right of $-1$ .

Figure 1 – The number line

For specific values, let’s choose $a = 2$ and $b = -1$ as shown in the diagram above and choose $c = -1$ . Note that we will just use these values for discussion purposes, but we may take any values. It would help, if we think of $a$ and $b$ as two points on the number line with $a$ as a blue point on the right $b$ , a red point.

And note that before multiplying with a negative number, VALUE OF BLUE POINT > VALUE OF RED POINT.

Since $a$ and $b$ are variables, we need to multiply all the numbers on the number line by $-1$ . This is to ensure that whatever values we choose for $a$ and $b$ , we multiply them by $-1$ . If we multiply every number on the number line by $-1$ , the geometric consequence would be a number line with negative numbers on the right hand side of $0$ , and positive numbers at the left hand side of $0$ as shown in Figure 2.

Figure 2 – Afer multiplying all numbers on the number line by -1

But negative numbers should be at the left hand side of $0$ so we reverse its position by rotating it 180 degrees from any point of rotation (for example, 0). The resulting figure is shown in Figure 3.

Figure 2 - All numbers in the number line were multiplied by -1

Notice that the blue and red points changed order and that the blue point is now at the left of the red point. Therefore, VALUE OF BLUE POINT < VALUE OF RED POINT. That is, why the inequality sign was reversed.

Summarizing, multiplying an inequality by a negative number is the same as reversing their order on the number line. That is, if $a, b$ and $c$ are real numbers, $a > b$ and $c<0$ , then $ac < bc$ .

Our summary above is actually a mathematical theorem. The proof of this is shown below. It is a very easy proof, so, I suppose, that you would be able to understand it.

Theorem: If $a, b$ and $c$ are real numbers, with $a > b$ and $c<0$ , then $ac < bc$ .

Proof:

Subtracting $b$ from both sides, we have $a - b>0$ .

Now, $a - b>0$ means $a - b$ is positive.

Since $c$ is negative, therefore, $c(a - b)$ is negative (negative multiplied by positive is negative)

Since $c(a - b)$ is negative, therefore, $c(a - b) < 0$ .

Distributing $c$ , we have $ac - bc < 0$ .

Adding $bc$ to both sides, we have $ac < bc$ which is what we want to show . $\blacksquare$

Leave a comment flip inequality sign, flip inequality sign dividing negative, inequality sign change, switch the inequality sign, when do you flip inequality sign

Problem Set 2

October 13, 2009 GB Problems Sets

PROBLEMS

1.) Find a linear function $f(x)$ such that $f(1) = 42$ and $f(2) = 47$ .

2.) Solve for $x$ : $4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 170.$

3.) Prove that the product of $3$ consecutive numbers is always divisible by $6$ .

4.) Prove that if $p$ is prime, $a$ and $b$ are integers, and $a \equiv b\mod p$ , then $a^p \equiv b^p \mod p$ .

SOLUTIONS AND PROOFS

Post Date: October 20, 2009

1. Solution: This is just the same as saying, find the equation of the line passing through $(1,42)$ and $(2,1337)$ . So, by point slope formula, we have, $y - y_1 = \displaystyle\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\Rightarrow y - 42 = \displaystyle\frac{1337 - 42}{2 - 1}(x - 1). \Rightarrow y = f(x) = -18x + 92.$

2.) Solution: $4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 4^x(1 + 4 +4^2 +4^3) = 170 \Rightarrow 4^x(85) = 170\Rightarrow 4^x = 2 \Rightarrow 2^{2x}=2^1 \Rightarrow x = \displaystyle\frac{1}{2}.$

3.) Proof: A number is divisible by $6$ if it is divisible by $2$ and $3$ . A product of $3$ consecutive numbers is divisible by $2$ because at least one of them is even, so it remains to show it is divisible by $3$ .

If a number is divided by $3$ , its possible remainders are $0, 1,$ and $2$ . Assume $n, n +1$ and $n+2$ be the three consecutive numbers, and $r$ be the remainder if $n$ is divided by $3$ .

Case 1: If $r=0$ , we are done.

Case 2: If $r = 1$ , then $n + 2 \Rightarrow r=0$

Case 3: If $r = 2$ , then $n + 1 \Rightarrow r = 0$ .

Since the product of the three consecutive numbers is even, and for each case of $r$ , one of the consecutive numbers is divisible by $3$ , the product of three consecutive numbers is divisible by $6. \blacksquare$

4.) Proof: From definition, $a^p \equiv b^p \mod p \Leftrightarrow b = a + kp$ for some $k \in \mathbb{Z}.$

Raising both sides of the equation to $p$ , we have $b^p = (a + kp)^p.$ By the binomial theorem, $b^p = (a + kp)^p = a^p + \displaystyle {p \choose 1}a^{p-1}kp + \displaystyle{p \choose 2}a^{p-2}k^2p^2 + \ldots + k^pp^p$ .

Notice that every term aside from $a^p$ is divisible by $p^2$ . (Why?). Therefore, $a^p \equiv 0 \mod p^2 .$

Hence, then $a^p \equiv b^p \mod p.$ $\blacksquare$

Leave a comment divisibility, exponents, linear function, modulo division, prime numbers

Problem Set 1

October 5, 2009 GB Problems Sets

PROBLEMS

1.) The sum of two numbers is $18$ and there difference is $-4$ . What are the two numbers?

2.) Find the values of $p, q$ and $r$ if:

$p + q + r = 3 \frac{1}{2}$
$pq + qr + rp = - 2 \frac{1}{2}$
$pqr = -2$

3.) Prove that $\displaystyle\frac{x + y}{2} \geq \sqrt{xy}$

4.) Define $cos(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$ and $sin(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$

a.) Prove that $cos(-x) = cos(x)$
b.) Prove that $sin(x) = -sin(x)$

SOLUTIONS AND PROOFS
Posted October 13, 2009

1.) Solution: Let $x$ and $y$ be the two numbers. Then, $x + y = 18$ and $x - y = -4$ . Adding the equations, we have $(x + y) + (x - y) = 18 + (-4) \Rightarrow 2x = 16 \Rightarrow x = 8$ . Substituting it to the first equation gives us $y = 10$ . Therefore, the two numbers are $6$ and $10$ .

2.) Solution: From the given, $p, q$ and $r$ are roots of of the cubic equation $x^3 - \frac{7x^2}{2} - \frac{5x}{2} + 2.$ Factoring, we have $(x+1)(x-4)(2x + 1)=0.$ Therefore, $x = 1, 4$ or $-1/2$

3.) Proof: We know that the square of the difference of any two numbers is always positive or $0$ . Let $x, y$ be any two numbers. Then, $(x - y)^2 \geq 0$ . Expanding, we have $x^2 - 2xy + y^2 \geq 0$ . Adding $4xy$ to both sides of the equation yields $x^2 + 2xy + y^2 \geq 4xy \Rightarrow (x + y)^2 \geq 4xy$ . Getting the square root of both, we have, $x + y \geq 2 \sqrt{xy} \Rightarrow \frac{x + y}{2} \geq \sqrt{xy}. \blacksquare$

4.) Proof (a): We want $\cos (-x)$ so we will just replace $x$ ‘s with $-x$ . Therefore, $cos(-x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-x)^{2n}}{(2n)!} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-1 \cdot x)^{2n}}{(2n)!} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-1)^{2n} \cdot x^{2n}}{(2n)!} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}= cos(x). \blacksquare$

Proof of 4b is left as an exercise. It’s very similar to the proof of 4a.