# Why do inequalities flip signs? [closed]

Is there a mathematical reason (like a proof) of why this happens? You can do it with examples and it is 'intuitive.' But the proof of why this happens is never shown in pedagogy, we just warn students to remember to flip the inequality when

• multiply or divide by a negative number both sides

$$-2>-3 \implies 2 < 3$$

• take reciprocals of same sign fractions both sides

$$\frac{3}{4} > \frac{1}{2} \implies \frac{4}{3} < 2$$

• – Nico A Nov 8 '18 at 2:47
• But the proof of why this happens is never shown in pedagogy --- I was curious about the book I used in high school algebra 1 (actually was the 1970 revised edition), since it was new-math based and heavy with proofs (actually, it was mostly heavy with terminological formalism), and it only gives some numerical examples of how multiplying by a negative number reverses the order and then simply states the results in a "Multiplication Axiom of Order" shaded section. – Dave L Renfro Nov 9 '18 at 8:29
• This will be useless for high school algebra students, but for those who already learnt differential calculus, it might be nice to note that this is essentially because the functions $f(x)=-x$ and $f(x)=1/x$ have derivatives that are negative everywhere. – YiFan Feb 6 '19 at 10:26
• It is appalling if your claim that "the proof of why this happens is never shown in pedagogy" is in fact true in general. After all, as mathematics teachers we ought to teach mathematics, not blind mimicking. – user21820 Sep 26 '19 at 8:54

## 9 Answers

I'm slightly concerned that

Is there a mathematical reason (like a proof) of why this happens?

is a purely mathematical question, but since you write "we just warn students" I will assume that this question is purposefully asked here on Math Educators StackExchange.

As to a proof:

Given $$a>b$$, subtract $$a$$ from both sides: $$0 > b-a$$.

Next, subtract $$b$$ from both sides: $$-b > -a$$.

Note that this final inequality is equivalent to $$-a < -b$$.

And so we have proved: If $$a > b$$, then $$-a < -b$$.

• Sweet, and the same works for reciprocation (if $a<b$ are of the same sign, divide by their strictly positive product $ab$) – Vandermonde Nov 8 '18 at 0:22
• thanks, I get that the concern for proof is technically mathematical but it isn't covered in any proof for any math courses since it's just elementary and assumed to just be understood (in the US anyways). I'm just curious how to show it if some student ever asks. – Lenny Nov 8 '18 at 4:45
• @Lenny This sort of thing is often proved in an introductory real analysis course. – Steven Gubkin Nov 8 '18 at 12:38
• The number of upvotes here is not proportional to the quality of my post. I hope those who are satisfied by this explanation will read my past answers, as well as answers from other users, which have generally involved significantly more work! – Benjamin Dickman Nov 10 '18 at 3:45

Depending on the context and the previous curriculum, the following might work:

1. "less than" means "to the left of" on the number line.
2. Multiplying by a negative number flips numbers around 0.
3. Thus, "left of" becomes "right of", or "greater than".
• This works for reciprocals because you flip (say) the positive reals around 1 (and squash all the numbers greater than one into the unit interval, and stretch the unit interval out to infinity). That's harder to show on a blackboard, but makes for a nice visual if you can find a way to demo it. – Kevin Nov 8 '18 at 3:22
• This seems like a visual translation of the intuitive understanding, but the question was asking for a mathematical explanation. – Barmar Nov 8 '18 at 20:46
• If inequalities are defined on the number line and multiplication by -1 as flipping, this is 100% mathematical. – Jasper Nov 8 '18 at 21:32
• I think this is a great answer, and would hope that students could use this method to provide a proof. It might be worth breaking down into cases, e.g., both negative, both positive, one negative and one positive; but, I certainly agree that this can constitute a "mathematical explanation"! – Benjamin Dickman Nov 14 '18 at 4:31

Perhaps one way to see (and explain intuitively to children) the "multiply by $$-1$$ part" is the following. Imagine your two numbers, $$a$$ and $$b$$, lying on the numberline. Multiplying by $$-1$$ is like 'rotating the numberline through 180°': imagine it's a straight metal pole, lying flat on the ground; pick it up by the middle, and rotate it 'long ways' (ie not 'barrel roll') 180°. (One only needs to consider the line segment $$[-\max\{a,b\},\max\{a,b\}]$$ for this, in case you get a smart-arse saying you can't move something of infinite mass!) It is hopefully clear to most people why the point that was to the right is now to the left.

This is only a visual intuition -- one of course needs to make this a rigorous proof, and the other answers do this -- however such intuition can often be valuable to early-learners.

A similar, but slightly less clean, statement can be made regarding inversion: take just the positive real axis $$(0,\infty)$$; rotate it around the number $$1$$. (Since, again, we can look at $$(0,M)$$ for say $$M = a + b$$ (with $$a,b > 0$$), one could imagine some sort of ellipse.)

I would draw graphics to illustrate... if only I were better at doing so!

PS. In complex analysis, multiplying by $$-1$$ really is rotating by 180°!

• [More generally, one can view a reflection in a dimension as being a rotation in the next dimension. E.g., a reflection over the origin here can be thought of as described above, but also a reflection of a point in 2D can be thought of as rotating that point through the third dimension. Anyway] I'm not sure (truly: unsure!) whether this explanation -- imagining a 'long ways' rotation of a metal pole -- would be helpful in building intuition for students with a nascent understanding of inequalities. – Benjamin Dickman Nov 14 '18 at 4:35
• My concern would be that reflections have determinant $-1$ while rotations $1$. While what you say is correct for the projection onto the lower dimensional space, you don't want to start confusion reflections/rotations, I feel. It is an interesting perspective, though, thanks for your comment! – Sam OT Nov 15 '18 at 12:21

For multiplying or dividing by -1... \begin{align} a&>b\\ a-b&>0\\ \\-(a-b)&<0\\ -a&<-b \end{align} (You can then extended to arbitrary negative numbers by multiplying or dividing by the [positive] magnitude.)

For taking reciprocals... assuming $$ab>0$$

\begin{align} a&>b\\ \left(\frac{1}{ab}\right)a&>\left(\frac{1}{ab}\right)b\\ \frac{1}{b}&>\frac{1}{a}\\ \frac{1}{a}&<\frac{1}{b}\\ \end{align}

• You might want to add some words in the slot where you flipped the sign from > 0 to < 0. My first reaction was that you just used the fact you were trying to prove. Anyway welcome to the site! – Chris Cunningham Nov 8 '18 at 13:29
• Thanks... when posting I tried (unsuccessfully) to use \intertext... so I tried a lineskip and went with that... avoiding further delay. Luckily(?), my post came in 2 minutes before the accepted answer by @Benjamin. – robphy Nov 8 '18 at 13:37
• @robphy What is the point of the third step? You can go straight from the second to the fourth by subtracting $a$ from both sides and it doesn't need an additional rule that $a > 0 \Rightarrow -a < 0$. – Reinstate Monica Nov 8 '18 at 19:42
• In the spirit of my second proof, I should have subtracted the sum (a+b) from the first line. But I wanted to explicitly do the "multiplication by -1" because that subtraction move might not be obvious to a novice. In the spirit of my first proof, I should have divided both sides by b, then taken the reciprocal explicitly so that a/b > 1 becomes b/a <1. Then divide both sides by b to get 1/a < 1/b. – robphy Nov 8 '18 at 22:18

In a more general case this happens if you deal with a strictly monotonic decreasing function like f(x) = -x or f(x) = 1/x (in the positive or negative numbers).

• Is monotonic monotonous? Just a humorous question. – JosephDoggie Nov 8 '18 at 16:14
• @ChrisCunningham Yes, you're right of course. That's what you get for expecting google to treat the term as an entity. – user10545 Nov 8 '18 at 17:34

I can't help but feel this question highlights the self defeating nature of teaching maths via "cheat tricks". In my opinion it is probably best not to teach students ever to just "switch the sign" of the inequality in an arbitrary manner, which can be confusing.

As elegantly illustrated by @Benjamin_Dickman's proof, the alternative notation $$a>b \quad \text{implies} \quad -b>-a$$ is much clearer, and invites the student to treat the inequality as a fixed equivalence in the same way they would treat an equality.

I feel it might be less confusing for students were this convention adopted, and the inequality was never reversed but for as a last resort. That would mean one less "rule" for students to "remember".

Which would you rather I give you: 10 cookies or 8 cookies? Okay, now which would you rather have me take away: 10 cookies, or 8 cookies? Adding 10 gives you a larger number than adding 8, but subtracting 10 means leaves you with a smaller number than subtracting 8 does. Similarly, multiplying by 3 gives you more than multiplying by 2, but dividing by 3 gives you less than dividing by 2.

• Moral plus one. – guest Nov 8 '18 at 19:03

I see it as a combination of the following, when we assume a field with a positive cone, and trichotomy. We want to show that if $$a and $$c<0$$, then $$cb.

• It is natural to expect the product of positive numbers to be positive.

• We denote the additive inverse of $$a$$ by $$-a$$.

• We note, from $$a+(-a)=0$$, that $$-(-a)=a$$.

• We note that $$(-a)b=-(ab)$$. Proof: $$(-a)b+ab=(-a+a)b=0\times b=0$$ (one can earlier deduce that multiplication by $$0$$ is $$0$$ from distributivity).

• Now $$(-a)(-b)=-(a(-b))=-((-b)a)=-(-ab)=ab$$.

• From the above we conclude that the product of two negative numbers is positive, and that a positive number times a negative is negative.

• Now if $$a and $$c<0$$, we have $$b-a>0$$ and $$-c>0$$, so $$(-c)(b-a)>0$$. That is, $$-cb+ac>0$$, or $$cb.

Other answers address multiplication by a negative number on both sides better than I can. For taking reciprocals of both sides, it may be helpful to examine why we know we can do that and maintain equality. This is one simple demonstration:

Starting with the equation $$\frac{a}{b}=\frac{c}{d}$$,

we can multiply both sides by $$\frac{d}{c}$$, giving $$\frac{ad}{bc}=\frac{cd}{dc}$$, which simplifies to $$\frac{ac}{bd}=1$$.

We can then do the same with $$\frac{b}{a}$$, giving $$\frac{bad}{abc}=\frac{b}{a}\cdot 1$$, which simplifies to $$\frac{d}{c}=\frac{b}{a}$$.

This shows that if two numbers are equal, then their reciprocals are equal. However, notice that we didn't just turn each number into its reciprocal. We also flipped their positions: the $$\frac{a}{b}$$ on the left became $$\frac{b}{a}$$, but on the right. When we're talking about equality, this is inconsequential, since equality is symmetric. However, if you take the same example with $$=$$ replaced with $$>$$ or $$<$$, the fact that the values "switched places" does matter.

This demonstration also shows why the sign only flips when both sides have the same sign. If they have different signs, then their reciprocals also have different signs, and exactly one of the multiplications by a reciprocal causes the inequality to flip. So you could argue that it does flip, but then it flips again, making it seem like it hasn't flipped. Similarly, if both sides are negative, it flips a total of three times, which has the same result as flipping once.