## Why do we reverse/flip the inequality sign?

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:

1. 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.
2. 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.

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 POINTVALUE 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 .$\blacksquare$

## Problem Set 2

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 $a^p \equiv b^p \mod p.$ $\blacksquare$

## Problem Set 1

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.

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