24
$\begingroup$

We are learning about inequalities. I originally assumed it would be the same as equations, except with a different sign. And so far, it has been - except for this.

Take the simple inequality: $-5m>25$ To solve it, we divide by $-5$ on both sides, as expected. $m>-5$.

But, I have been told that now we have to flip the inequality sign because we divided by a negative (and this also applies to multiplying negatives).

$m<-5$

And this does work. Plug in any value less than $-5$ and it does turn out to be more than 25, but why?

Mathematically, why do we flip the sign here?

$\endgroup$
  • 16
    $\begingroup$ Not a proof and using casual language, you can think about this. $15>0$ Multiplying each side by $-1$ gives $-15>0$ but you know this isn't true. The side that used to be more 'positive' is now more 'negative'. Also, imagine both sides of the inequality are positive. Then the greater side is the one with a higher absolute value. Multiplying both sides by -1 will make the side that had more (the greater side) absolute value, now be the side with a more 'negative' number and thus be the lesser side. You can similarly work through all the possible combinations of negative and positive values $\endgroup$ – turkeyhundt Nov 24 '15 at 1:47
84
$\begingroup$

Surely you believe that we can add/subtract from inequalities without a problem. I show you why using this.

If you have that $x>y$, then subtract $y$ to get $x-y>0$ and subtract $x$ to get $-y>-x$. That is, multiplying by $-1$ flips the inequality.

$\endgroup$
  • 9
    $\begingroup$ Simple and elegant explanation. $\endgroup$ – Ruslan Nov 24 '15 at 12:32
18
$\begingroup$

The act of multiplying by a positive scalar is to stretch the number line outward from the origin (or shrink inward if the scaling factor is less than one). If one point on the number line is to the left of another, that fact remains true after stretching. Multiplying by a negative not only stretches/shrinks it but also flips it across the origin - think of it as a $180^\circ$ rotation. If you do that to two points, then that will flip what order they were in. If point A was to the left of point B to begin with, then after flipping, point B will be to the left of A afterwards.

In symbols, $a<b\implies ra<rb$ if $r>0$ and $a<b\implies rb<ra$ if $r<0$.

One may prove the above axiomatically, using nothing but addition and subtraction as avid19 says.

$\endgroup$
  • $\begingroup$ @TreFox: I recommend this as the intuitive answer, because it is the most intuitive way to correctly view multiplication of real numbers, and because it extends naturally to complex numbers, where multiplying by $i$ corresponds to a $90^\circ$ rotation, which makes the relation $i^2 = -1$ intuitively satisfying. The other answer by avid19 is perfectly fine if you only care about following the field axioms, and even then you've to first prove that $-1 \times x = -x$, otherwise his answer is incomplete. $\endgroup$ – user21820 Nov 24 '15 at 14:04
  • 2
    $\begingroup$ @user21820 If you're at the level of "algebra-precalculus" as the tag indicates, you're not going to be aware of field axioms, nor do you need to worry about proving that -1x = x. $\endgroup$ – Matt Gutting Nov 24 '15 at 19:44
  • $\begingroup$ @MattGutting: Exactly my point, which means that the other answer is simply not accessible to precalculus level unless they want to learn the field axioms, otherwise it is begging the question since the core part of the fact is actually $-1 \times x = -x$! $\endgroup$ – user21820 Nov 25 '15 at 5:50
  • $\begingroup$ @user21820 I don't think $-1\times x=-x$ is inaccessible to precalculus level. As a matter of fact I think it is well known and understood by most people at that level. $\endgroup$ – user223391 Nov 25 '15 at 16:47
  • 1
    $\begingroup$ @avid19 To be fair, the overwhelming majority of people that know $-1\cdot x=-x$ do not know how to prove it (from the distributive property), may not even have any familiarity or exposure to proofs in math at all, and may not even know that $-x$ is by definition the additive inverse of $x$. $\endgroup$ – whacka Nov 25 '15 at 21:48
9
$\begingroup$

For those who benefit from imagery, consider this number line:

$<----- (-5) ---- 0 ------ (7) --->$

You can see $7$ is farther to the right than $-5$, so $7 > -5$.

Multiply both of those values by $-1$, and you flip the number line:

$<--- (-7) ------ 0 ---- (5) ----->$

Now you can see $5$ is farther to the right than $-7$, so $5 > -7$, or $-7 < 5$.

You can extend that logic to an equation with variables, like the example in the question:

$-5m>25$

is represented on the number line as follows:

$<------------- 0 ----- (25) ----- (-5m) --->$

Divide both sides by $-5$ to get $m$ by itself. As before, this flips the number line. It also scales the whole thing down by a factor of 5. The result:

$<------- (m) - (-5) - 0 ------------------>$

That number line can be represented as $-5>m$ or, as the question points out:

$m < -5$

$\endgroup$
3
$\begingroup$

Imagine two points on a number axis, say $1$ and $3$. Certainly $1$ is to the left of $3$, which we write $$1<3$$ Now, let's multiply both sides by $2$. That means we scale the situation to a twice bigger: $1$ becomes $2$ and $3$ lands at $6$. Of course what was on the left in the pair, is still on the left: $$2<6$$

Multiplying by a negative value, however, is not just scaling, it is also flipping the whole image with respect to $0$ (zero). It is like you have rotated the line $180^\circ$ around a pivot at the origin. That way what was previously on the left side now appears on the right side: $$(-6)<(-2)$$

The values actually get swapped, but we can equivalently keep them on their previous sides and flip the inequality direction instead: $$(-2)>(-6)$$

$\endgroup$
2
$\begingroup$

It's just a matter of equation reforming.

If you have $x < y$ this is equivalent to $f(x) < f(y)$ if and only if $f$ is a strictly monotonically increasing function. If $f$ is a strictly monotonically decreasing function (like multiplying with a negative number is), it flips the inequality. If the function is only non-decreasing, you get $f(x) \le f(y)$ and cannot get back to the original equation any more. If $f$ has both increasing and decreasing parts (like $f(x)=x^2$), you need to split the (in)equality into monotonically increasing and decreasing parts and combine the results.

$\endgroup$
  • $\begingroup$ Your last sentence doesn't make any sense. If $x < y$ how are you intending to compare $x^2$ and $y^2$? $\endgroup$ – user21820 Nov 24 '15 at 13:58
  • $\begingroup$ @user21820 makes good sense to me. i.e. given $x < y$, $x^2 ? y^2$ splits into $\left\{\begin{array}{11} x^2 < y^2 & x > 0, y > 0 \\ x^2 > y^2 & x < 0, y < 0 \end{array} \right. $ $\endgroup$ – hobbs Nov 25 '15 at 15:50
  • $\begingroup$ @hobbs: Nonsense. You missed out $x \le 0 \le y$ which is precisely where nothing can be said about the order between $x^2,y^2$. $\endgroup$ – user21820 Nov 26 '15 at 2:56
  • $\begingroup$ @user21820 I'm well aware of that. I said nothing because nothing can be said. That doesn't make anything "nonsense". $\endgroup$ – hobbs Nov 26 '15 at 3:27
  • $\begingroup$ (not saying anything in cases where there's nothing meaningful to be said is an example more people should follow) $\endgroup$ – hobbs Nov 26 '15 at 3:31
1
$\begingroup$

It is just the fact that the product of a negative number by a positive is negative.

You have $25-5m>0$. If you multiply by the negative number $-1/5$, the result is negative: $$ (-1/5)(25-5m)<0. $$ After expanding, $$ m-5 <0. $$

$\endgroup$
1
$\begingroup$

One way to look at this problem, is to change the problem to one with a positive coefficient.

\begin{array}{lrcl} \text{We start with : } & -5m &\gt &25\\ \text{Add $5m$ to both sides : } & 0 &\gt &5m + 25\\ \text{Add $-25$ to both sides : } & -25 &\gt &5m\\ \text{Make $5m$ the subject : } & 5m &\lt &-25\\ \text{Multiply both sides by $\frac 15$ :} & m &\gt &-5\\ \end{array}

Once you have worked out the logic of this solution, it makes more sense that the direction of the inequality changes when you multiply both sides by a negative number.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.