MHB Why Does an Inequality Sign Flip?

[emacat]

Why would an inequality sign flip in an answer.
For example:
16 < -s -6
The answer is given s < -22
I had -s > 22

I am thinking it is because when you x by -1 to keep from having a negative variable the inequality sign flips...is this why?

 

[MarkFL]

If we are given:

$$16<-s-6$$

We can then add $s-16$ to both sides to get:

$$s<-22$$

Your result is equivalent to this, but we typically want to solve for the variable itself, and not its negative, and so yes you are correct, if you multiply an inequality by a negative value, the direction of the inequality must be reversed.

 

[samir]

Hi!

I understand the question but I am not sure how to answer it in a way that helps you build an intuition for it. But I will show you the mechanics.

First method

$16<-s-6$

$16+6<-s-6+6$

$22<-s$

The moment you multiply the negative variable by -1 you flip the inequality sign.

$22 \cdot (-1) > -s \cdot (-1)$

$-22 > s$

Notice how I flipped the sign before I simplified both sides. The statement where you multiply by -1 is where you flip the sign. Think of this if you like to rewrite your statements many times and do things thoroughly and step by step. The inequality statement you had previously is no longer the same if you multiply both sides by -1.

$s < -22$

I only flipped the sides here so we can easily compare it to the correct answer.

Second method

This second method is really the same method, but I divide by -1 instead of multiplying by -1.

$22<-s$

$\frac{22}{-1}>\frac{-s}{-1}$

Same thing here, I flip the sign as soon as I write down the division by -1, before even simplifying it.

$-22>s$

$s<-22$

Third method

This is the way Mark explained it.

$16<-s-6$

Make note of the following.

$s$ is the inverse of $-s$

$-16$ is the inverse of $16$

Add the sum of the inverses (s+(-16)) to both sides of the inequality.

$16+(s+(-16))<-s-6+(s+(-16))$

Simplify the sum of the inverses.

$16+(s-16)<-s-6+(s-16)$

Remove the parentheses.

$16+s-16<-s-6+s-16$

Commute the terms with equal absolute value.

$16-16+s<-s+s-6-16$

$s<-22$

Why...

I am not sure how to explain the "why" part, why we flip the sign. But inequalities essentially compare two quantities, where one is of lesser quantity, or lesser value than the other.

 

[Deveno]

Here is the "why" about the "sign flipping" (have patience with me, sometimes I don't explain things so well).

If we write:

$a < b$, then subtracting the same thing from both sides, still should preserve the "gap" between them:

$a - a < b - a$ (now, instead of comparing $a$ to $b$, we've "shifted" both amounts by $-a$...if $a$ was originally positive, this is $a$ units "to the left").

But, as anyone should know, $a - a = 0$, so we see that if:

$a < b$, then $0 < b - a$. We can run this argument backwards, if we start with:

$0 < b - a$, then surely $a < (b - a) + a$, so that: $a < (b + (-a)) + a = b + ((-a) + a) = b + 0 = b$.

So we could actually use this as a DEFINITION of "$<$":

$a$ is less than $b$, if $b - a$ is positive.

We could, in a similar fashion, go on to say that $a = b$ if $b - a = 0$, and $a > b$ if $b - a < 0$.

So let's say we KNOW, that $a < b$. What is the relationship between $-a$ and $-b$?

To find out, we need to look at the sign of $-b - (-a)$.

Now $-b - (-a) = - b + a = a - b = (-1)(b - a)$.

It may seem "obvious" that if $b - a > 0$, that $(-1)(b - a) < 0$, although to prove this, I would have to digress even more. I urge you to think about this, though.

Now, since $-b - (-a) = (-1)(b-a) < 0$, by our definition above, we conclude that $-a > -b$.

***********

It helps to keep the following examples in mind:

$3 < 4$, so $-4 < -3$ (this is the same as writing $-3 > -4$).

$-2 < 2$, so $-(-2) = 2 > -2$.

$-2 < -1$, so $-(-2) = 2 > 1 = -(-1)$.

 

[samir]

Let me try to explain the "why" using some numeric patterns and examples.

Consider this pattern.

$$-4=-4$$
$$-3=-3$$
$$-2=-2$$
$$-1=-1$$
$$0=0$$
$$1=1$$
$$2=2$$
$$3=3$$
$$4=4$$

Now consider this pattern.

$$-4<4$$
$$-4<3$$
$$-4<2$$
$$-4<1$$
$$-4<0$$
$$-4<-1$$
$$-4<-2$$
$$-4<-3$$
$$-4=-4$$
$$-4>-5$$
$$-4>-6$$

Notice how the sign flipped at $-4>-5$. Also, note the point at which the two sides are equal.

Now consider the following.

$-8<12$

$-8 \cdot -1 > 12 \cdot -1$

$8 > -12$

Notice how the sign flipped when both sides were multiplied by $-1$. It should be clear why this is. Negative number times a negative is positive. Positive times a negative is a negative. A positive number cannot be less than a negative number. The opposite is also true, a negative number cannot be greater than a positive number.

Now consider this statement.

$16<-s-6$

This says that $-s-6$ is greater than $16$. It also says that $16$ is less than $-s-6$. How can that be? The $16$ is a positive integer, while both $-s$ and $-6$ are negative. How or when is $-s-6$ greater than $16$? Let's find the point at which $-s-6$ is equal to $16$ first.

$-s-6=16$

Solve the equation for $s$.

$-s-6+6=16+6$

$-s=22$

$-s \cdot -1 = 22 \cdot -1$

$s=-22$

Notice! When $s=-22$ the $-s-6$ is equal to $16$.

$16<-s-6$

$s=-22$

$16=-(-22)-6$

$16=22-6$

$16=16$

So what would it take to make the 16 on the left side less than the number on the right of the equal sign? For what value of $s$ is $-s-6$ greater than $16$? We know that $s=-22$ makes them equal. So if we add $-1$ to that, it should make them not equal.

$-22+(-1)=-22-1=-23$

$16<-s-6$

$s=-23$

Plug this value in on the right side of the inequality and evaluate.

$-(-23)-6=23-6=17$

$16<17$

Consider the original statement again.

$16<-s-6$

Isolate the negative s.

$16+6<-s-6+6$

$22<-s$

The only way for $-s$ to be greater than the positive integer $22$ is if we plug in a negative value for $s$, or if the absolute value of $-s$ is greater than $22$.

$22<-s$ where $|-s|>22$

$|-s|>22$ when $s<0$

I hope that helps the poster make some additional connections.