$(1+x)(1-x)=1-x^2$

Question

If a number increases by $x\%$ and then decreases by $x\%$ it will be decreased by $\dfrac{x^2}{100}\%$ since $(1+x)(1-x)=1-x^2$. I understand that the decrease is on a value greater than the initial value. My question is why is there an asymmetry between addition and subtraction (or between increasing and decreasing) with respect to multiplication?

To clarify a little bit more:
$(+x)(+x)=(-x)(-x)=+x^2$ and $(+x)(-x)=(-x)(+x)=-x^2$ so there a $50\%$ chance of getting a negative result; the situation is symmetric.
increase followed by increase: increase
increase followed by decrease of same percentage: decrease
decrease followed by increase of same percentage: decrease
decrease followed by decrease: decrease
so there's a $75\%$ chance of the result being a decrease. Why is there an asymmetry in this case?

Answer 1

My first reaction was that yes, there is asymmetry in the sense you are speaking of. It seems that if you denote by $I$ increase by percentage and by $D$ decrease by percentage, that the whole thing corresponds to multiplication table

\begin{array}{c|c|c} & I & D\\ \hline I & I & D\\ \hline D & D & D \end{array}

(This is just multiplication in $\mathbb Z/2\mathbb Z$, which arises naturally with many binary properties. For example, consider even and odd numbers, adding them corresponds to addition in $\mathbb Z/2\mathbb Z$, while multiplying them to multiplication in $\mathbb Z/2\mathbb Z$ - you get the same "asymmetry".)

However, this is not an accurate description. If you fix some percentage, then you just don't have $I^2 = I$, or any other of the identities from the above table, simply for the reason that increase/decrease is by some different percentage. If you want to say, that $I$ is any increase and $D$ is any decrease, then $ID$ is undefined.

So, instead of fixing a percentage, let us just consider $(-1,\infty)\subseteq \mathbb R$ and binary operation $x*y = x+y+xy$ on it. It turns out that this is a group.

But, why this operation?

Well, it turns out that $(1+x)(1+y) = 1+ x+y+xy = 1+ x*y$, so this group models our increase/decrease by percentage. Negative numbers correspond to decrease and positive to increase.

The neutral for this operation is $0$ and the inverse is given by $x^{-1} = \frac{-x}{1+x}$. If $x\in(-1,0)$, then $x^{-1}>0$ and $|x^{-1}| > |x|$, which we interpret as decrease "being stronger" than increase.

If we extend this operation to $[-1,\infty)$, it is no longer group, but we don't actually care about it. Pick any probability distribution $D$ on $[-1,\infty)$ and consider two independent equally distributed random variables $X,Y\sim D$. Let $\mu = E(X) = E(Y)$ be the expected value. Then, $E(X*Y) = E(X+Y+XY) = E(X)+E(Y) + E(X)E(Y) = 2\mu + \mu^2$.

Now, this is where it gets interesting, $\mu^2 + 2\mu < 0$ if and only if $\mu < 0 $, $\mu^2 + 2\mu >0$ if and only if $\mu > 0 $ and $\mu^2 + 2\mu = 0$ if and only if $\mu = 0$. There is no asymmetry: if you pick random increase/decrease in percentage, what you can expect depends on distribution you chose. Unless you were specifically biased towards choosing decrease, you shouldn't expect to get overall decrease.

TL;DR There is no asymmetry.

Answer 2

Well, you start from a different quantity the second time (when you do the decrease). That is why. So actually it would be asymmetrical if there was no asymmetry.

Answer 3

There's no dissymmetry between addition and subtraction: you do not add whatever fixed quantity when you increase in percentage. Or if I follow your reasoning, when you apply twice an increase of x %, you should obtain an increase of 2x %, which is not the case — simply because addition is not multiplication.

Answer 4

% increase or decrease has a larger effect when it's working on larger numbers. So, there's two possibilities here:

You increase first, then decrease. The decrease is working with the larger number, so the decrease is greater than the increase.

You decrease first, then increase. Once again, the decrease is working with the larger number, so the decrease is greater.

In all cases, a % decrease is slightly stronger than a % increase.

Answer 5

There's no asymmetry : $(1+x)(1-x) = (1-x)(1+x) = 1-x^2$.

Since $0\leq x \leq 1$, multiplying any number by $1-x^2$ will result in a decrease, since $0\leq 1-x^2 \leq 1$

Edit - after seeing yours- :

$(1-x)(1-x) = x^2 -2x +1$ ; and for $x\in [0;1]$, $0\leq x^2-2x+1 \leq 1$. Thus, applying two decreases will result in a decrease.

This is not comparable with $(-x)(-x)=x^2$, because applying a decrease of $\frac{t}{100}$ is not the same thing than multypling by $-\frac{t}{100}$. You're comparing two completely different operations.

Answer 6

This is because $\%$ is somewhat 'biased' in that it 'favours' a decrease, due to associativity of multiplication. Here is what it means:

Increase, Decrease

The number $a$ increases by $x\%$: $$a(1+0.01x)$$ This number $a(1+0.01x)$ decreases by $x\%$: $$a(1+0.01x)(1-0.01x)=a(1-0.0001x^2)<a$$

Decrease, Increase

The number $a$ decreases by $x\%$: $$a(1-0.01x)$$ This number $a(1-0.01x)$ increases by $x\%$: $$a(1-0.01x)(1+0.01x)=a(1-0.0001x^2)<a$$

You can't compare the 'probability' of $75\%$ to positives and negatives since we aren't even multiplying by a negative. We are multiplying something slighter greater than $1$ by something slightly smaller than $1$, to give something slightly smaller than $1$.

In short, you couldn't really compare a decrease in a number to a change in sign.

Answer 7

It should be expected, but unfortunately it frequently isn't because people don't realize that multiplication is involved.

If you increase an amount by $20\%$ and then again by $20\%$, then the total increase will be $44\%$, not $20\%$. If the initial price is $\$100$, the price after the first increase will be $\$120$; the next increase will add $20\%$ of $\$120$, that is, $\$24$. The end price will be $\$144$. Surprised? Perhaps not you, but some people would be.

Now, the same, but with a price drop by $20\%$ after an increase of $20\%$: the $\$24$ must be subtracted, ending up at $\$96$.

It's all in that $20\%$ (or, in general, $x\%$) of an amount is larger than $20\%$ of a smaller amount.

When you do “increase by $x\%$ and decrease by $x\%$” you always end up with a smaller amount than you started with.

If you instead do “decrease by $x\%$ and increase by $x\%$” you end up again with a smaller amount: nothing mysterious, because the result is exactly the same as before, that is, the initial amount is multiplied by $1-(x/100)^2<1$.

Decrease $\$100$ by $20\%$; you end up at $\$80$. Now increase by $20\%$, that is, $\$16$: you end up at $\$96$.

Nothing more than realizing that $$ a \underbrace{\Bigl(1+\frac{x}{100}\Bigr)}_{\text{increase by $x\%$}} \,\underbrace{\Bigl(1-\frac{x}{100}\Bigr)}_{\text{decrease by $x\%$}} = a \underbrace{\Bigl(1-\frac{x}{100}\Bigr)}_{\text{decrease by $x\%$}} \,\underbrace{\Bigl(1+\frac{x}{100}\Bigr)}_{\text{increase by $x\%$}} $$

Answer 8

To begin with, let's be clear about what $x$ is. Is it a percentage, or is it just a number? If it is a percentage, then the action of increasing by $x\%$ and then decreasing by $x\%$ is not represented by $(1+x)(1-x)$; rather, it is represented by $\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right).$ So I'll suppose you did not really mean $x\%,$ you meant $100x\%,$ so that when you divide by $100\%$ (in order to convert the percentage to a simple number) you get $x.$

Assuming that's what you meant, I wonder if part of the confusion is that the product $(1+x)(1-x) = 1 - x^2$ looks simpler than it really is. You look at the product $(+x)(-x) = -x^2$ and (apparently) see how it relates to the term $-x^2$ in $1 - x^2,$ and you probably understand how the $1$ in $1 - x^2$ relates to the two $1$ terms in $1 + x$ and $1 - x,$ but that's still only half of the picture.

In general, when you take two quantities produced by addition or subtraction and multiply them together, the multiplication distributes over the addition like this: $$ (a + b)(c + d) = ac + bc + ad + bd. $$ The product $(1+x)(1-x)$ is just a particular example of this operation, in which we set $a = 1,$ $b = x,$ $c = 1,$ and $d = -x.$ The result is $$ (1+x)(1-x) = 1\cdot 1 + (+x)\cdot 1 + 1\cdot(-x) + (+x)(-x) = 1 + x - x - x^2. $$ Conveniently, the two terms $x$ and $-x$ add up to $0,$ and we're left with just $1 - x^2.$ But for some other combination of signs for the $x$ terms in $(1+x)(1-x),$ you get a more complicated result Let's work out all four possibilities: \begin{align} (1 + x)(1 + x) &= 1 + x + x + x^2 = 1 + 2x + x^2, \\ (1 - x)(1 - x) &= 1 - x - x + x^2 = 1 - 2x + x^2, \\ (1 + x)(1 - x) &= 1 + x - x - x^2 = 1 - x^2, \\ (1 - x)(1 + x) &= 1 - x + x - x^2 = 1 - x^2. \end{align}

If $x$ is the amount of a typical increase from a typical percentage problem, then $0 < x < 1,$ and so $x^2 < x$ and $x^2 - 2x < 0.$ That's why the second possibility, $(1 - x)(1 - x)$, produces a result less than $1,$ that is, a decrease, instead of the increase that you would expect if you look only at the product $(-x)(-x).$

But I think there's also a deeper conceptual issue here:

... why is there an asymmetry between addition and subtraction (or between increasing and decreasing) with respect to multiplication?

Multiplication is not inherently asymmetric with respect to increasing and decreasing. The apparent asymmetry in the percentage increase and decrease arises because multiplication increases or decreases things by ratios while addition increases or decreases things by differences; and while the opposite of a given difference can be found by taking the negative of the difference, the opposite of a given ratio is the reciprocal of that ratio.

So, for example, we can increase something by multiplying by $\frac54.$ The reciprocal of $\frac54$ is $\frac45,$ so to undo the multiplication by $\frac54$ we multiply by $\frac45.$ When we do that, we end up exactly where we started.

But multiplication and ratios also seem to be less comfortable for people to deal with than addition and differences, so instead of simply taking a quantity multiplied by $\frac54,$ we take $\frac14$ ($25\%$) of the quantity added to the original quantity. Describing things in terms of percentage increases and decreases is mathematically more complicated than just multiplying by ratios, but somehow in many contexts we seem to prefer to say "$25\%$ increase" rather than "$\frac54$ as much" and "$20\%$ decrease" rather than "$\frac45$ as much," making a fundamentally symmetric pair of operations appear asymmetric.