Xtykutl

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".

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 $ab>0$

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

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°!

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).

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.

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".

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 $a<b$ and $c<0$, then $cb<ac$.

• 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: $(-a)b+ab=(-a+a)b=0times b=0$ (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 $a<b$ and $c<0$, we have $b-a>0$ and $-c>0$, so $(-c)(b-a)>0$. That is, $-cb+ac>0$, or $cb<ac$.

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.