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To increase a number by a percentage I was taught the formula no x 1.percentage, so for example:

100 increased by 15%:
100 x 1.15 = 115

But today, I learned the hard way that doing the inverse of the above (i.e. no / 1.percentage), isn't the correct way of decreasing a number by a percentage (aka "applying a discount"). E.g:

100 decreased by 15%:
100 / 1.15 = 86.95652 <- WRONG
100 - (100 x 0.15) = 85 <- RIGHT

My Question: Why doesn't the formula no / 1.percentage work, when decreasing a number by a percentage?

Side Question: Is there a simpler formula to decrease a number by a percentage than no - (no x 0.percent)?

Note: My math knowledge is super basic, so maybe pretend you're explaining this to a twelve year old when answering. But if you want to give sophisticated answers for future readers, that's fine too, I might just not understand it though.

The method of increasing a value $V$ by $p%$ is actually adding $Vtimesfrac p100$:
$$V_textnew = V + Vtimesfrac p100 = Vtimesleft(1 + frac p100right)$$
For $p=15$ you have a nice multiplier
$$left(1 + frac p100right) = 1 + 0.15 = 1.15$$

Remember, however, that '1.p' is a shortcut or mnemonic, not a method. It wouln't even work for $p$ greater than 99, say for $+120%$. And it won't work for $p <0$, either.

For decreasing you need to apply the method, which is:
$$V_textnew=V - Vtimesfrac p100 = Vtimesleft(1 - frac p100right)$$
so for 15-percent decrement you get:
$$V_textnew=V - Vtimesfrac 15100 = Vtimes(1 - 0.15) = Vtimes 0.85$$

EDIT

To answer your main question directly, forget percentages. Suppose you need to rise a value from 100 to 125. The new value is $$125 = 100timesfrac 54=100times( 1+frac 14) $$ hence an increase by 25%.
Now, if you want to reduce it back to 100, you get $$100=125times frac 45=120times(1-frac 15)$$ which is 20% decrease.
Why was it 1/4 before, and 1/5 now? Because the same difference $pm 25$ was taken relative to 100 in the former case and to 125 in the latter one. When we added a fourth part, we got five fourths of the initial value, so we needed to take away one of those five, i.e. a fifth part to get back.

And here you have the difference in percentages: 1/4=25%, while 1/5=20%.

So, $value/1.p$ division reduces the value by p% of the resulting value, not of the original one.

The reason that your suggested formula doesn't work is that "increase by $k$%" and "decrease by $k$%" do not undo one another like $+5$ and $-5$ or $times 2$ and $div 2$ do (we would say that they are not inverses).

For example $100$ increased by $50%$ is $150$. However, $150$ decreased by $50%$ is 75.

Therefore, if we want to decrease a number by a percentage, we can't just attempt to do the opposite operation of increasing by that percentage.

What you did is the right answer to a different problem. The general method, "multiply by $1+$ change" is a good one when used properly.

To increase a number by $15%$ you multiply by
$$
1 + 0.15 = 1.15.
$$
To decrease a number by $15%$ you multiply by
$$
1 - 0.15 = 0.85.
$$
But to undo a $15%$ increase you have to "unmultiply" by $1.15$. So you divide by $1.15$. Since
$$
frac11.15 = 0.86956521739 approx 0.87 = 1 - 0.13,
$$
to undo a $15%$ increase you make a $13%$ decrease.

To find $15 %$ of some number $N$, we solve an equation (proportion) $fracN100=fracx15$ or $x=0.15N$. That's why a $15%$ increase can be found as $N+0.15N=1.15N$ and decrease is found as $N-0.15N=0.85N$

It works the same way...

Increase : to increase of $15 %$ means to multiply by $(1+0,15)=1,15$.

Decrease : to decrease of $15 %$ means to multiply by $(1-0,15)=0,85$.

Unfortunately, the $15 %$ factors get similar results; thus it can be misleading.

Consider instead an increase of $20 %$. Why the "inverse" operation (i.e. to divide by $1,2$ to decrease does not produce the correct result ?

With a $20 %$ increase the updated price will be $120$.

Now, what is the amount of a decrease of the updated price by $20 %$ ? It is not $20$ but $24$.

And we have that $120 times (1-0,2)=96$ while $dfrac 1201,2=100$.

Consider the literal meaning of "per cent"

Increase $X$ by 15%:
$$
X + X times frac15100 = frac100X100 + frac15X100 = frac(100+15)X100 = frac115X100 = 1.15X
$$

Decrease X by 15%:
$$
X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
$$

If we want to increase $n$ by $p$%, we work out what $p$% of $n$ is, and then add it to $n$ itself: $$0.ptimes n+n=(0.p+1)times n=1.ptimes n$$This is how you arrived to your formula.

If we want to decrease $n$ by $p$%, we work out what $p$% of $n$ is, and then subtract it from $n$ itself:$$n-0.ptimes n=(1-0.p)times n$$This is how to derive the formula for a decrease.