Calculating discount not working 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.
 A: 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
$$
\frac{1}{1.15} = 0.86956521739 \approx 0.87 = 1 - 0.13,
$$
to undo a $15\%$ increase you make a $13\%$ decrease.
A: 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 $\dfrac {120}{1,2}=100$.
A: To find $15 \%$ of some number $N$, we solve an equation (proportion) $\frac{N}{100}=\frac{x}{15}$ or $x=0.15N$. That's why a $15%$ increase can be found as $N+0.15N=1.15N$ and decrease is found as $N-0.15N=0.85N$
A: The method of increasing a value $V$ by $p\%$ is actually adding $V\times\frac p{100}$:
$$V_\text{new} = V + V\times\frac p{100} = V\times\left(1 + \frac p{100}\right)$$
For $p=15$ you have a nice multiplier
$$\left(1 + \frac p{100}\right) = 1 + 0.15 = 1.15$$
Remember, however, that '1.p' is a shortcut or mnemonic, not a method. It wouldn't even work for $p$ greater than 99, say for $+120\%$. And it certainly won't work for $p <0$.
For decreasing you need to apply the method, which is:
$$V_\text{new}=V - V\times\frac p{100} = V\times\left(1 - \frac p{100}\right)$$
so for 15-percent decrement you get:
$$V_\text{new}=V - V\times\frac {15}{100} = V\times(1 - 0.15) = V\times 0.85$$
EDIT
To answer your main question directly, forget percentages. Suppose you need to rise a value from 100 to 125. The new value is $$125 = 100\times\frac 54=100\times(1+\tfrac 14) $$ hence an increase by 25%.
Now,  if you want to reduce it back to 100, you get $$100=125\times \frac 45=120\times(1-\tfrac 15)$$ 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, a division like $value/1.p$ reduces the value by p% of the resulting value, not of the original one.
A: 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.
A: If we want to increase $n$ by $p$%, we work out what $p$% of $n$ is, and then add it to $n$ itself: $$0.p\times n+n=(0.p+1)\times n=1.p\times n$$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:$$n-0.p\times n=(1-0.p)\times n$$This is how to derive the formula for a decrease.
A: Consider the literal meaning of "per cent"
Increase $X$ by 15%:
$$
X + X \times \frac{15}{100} = \frac{100X}{100} + \frac{15X}{100} = \frac{(100+15)X}{100} = \frac{115X}{100} = 1.15X
$$
Decrease X by 15%:
$$
X - X \times \frac{15}{100} = \frac{100X}{100} - \frac{15X}{100} = \frac{(100-15)X}{100} = \frac{85X}{100} = 0.85X
$$
