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original post

the examples here are, the most important word -- fundamentally -- the same.

  • example1: the most abstract way to present this example.

    Why equivalent % increase of A in event1 and % decrease of B in event2 is not the same end result in these two cases/events? % is percent chance of curing cancer. that logically means A = [total cancer potency], and B = [whatever the trade is for increasing the chance of curing cancer] ..and the "naming" of A and B is defined by choice

  • example2: concrete

    1 cupcake = 1 karma

    case/event 1

    • 66% increase of 100 cupcakes = .66 * 100 cupcakes = 66 increase, so total 166 cupcakes
      so the outcome is 166/100 = 1.66 cupcakes per karma

    case/event 2

    • 66% decrease of 100 karma = .66 * 100 karma = 66 decrease, so 34 total karma
      so the outcome is 100/34 = 2.9 cupcakes per karma

so to me and everyone, 66% increase and 66% decrease seems like they would lead to the same end result/outcome, BUT as shown by math calculation, the resulting outcome of each event is not the same. you are actually getting more cupcakes if you decrease the karma than you would if you increase the amount of cupcakes, which doesnt make any sense at all. you think to yourself how could this have happen? how can this be? the world doesnt make any sense. see talk for more.

looking for a non-math-calculation explanation, (cogintive, neuroscientific, lingiustical, philosophical, whatever is useful, etc.) using plain english, not math language since i was never taught it properly or clearly

==
a related question after thinking about the talk, with some deleted comments, below -- isnt it true that equal % increase of A is going to have the same end result as an equal % decrease in A if they have the same initial number?

  • if true, ok, so this would support the thinking in the original post
  • if false, i need to correct my assumptions, and yet, nobody has yet to explain clearly and helpfully why this is

i think this is true because they are equal % from the same initial number.

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umm.. if i use 50% to think about this, it really didnt help, 50% increase means you get 50% (half) more, 50% decrease means you get 50% (half) less –  kittensatplay Oct 1 '12 at 0:35
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In the scenario when you go half up and then go half down, going half down takes you down by more than going half up takes you up. –  alex.jordan Oct 1 '12 at 0:38
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"50%" is not a fixed amount, like 100 dollars. "50%" has to be 50% of something. When you go up 50%, and then down 50%, you are not taking 50% of the same thing each time, so you will not end up at the same place you started. –  Ted Oct 1 '12 at 0:41
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@high if they have the same starting points, then they get the same amounts. A $50\%$ increase from $100$ and a $50\%$ decrease from $100$ both give a difference of $50\$$. If that wasn't your complaint, what are you referring to? –  Robert Mastragostino Oct 1 '12 at 0:55
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If you don't think that "increase" and "decrease" mean the same thing by themselves, why would "50% increase" and "50% decrease" mean the same thing? I don't even know why you're so confused. There is no "math world" vs "real world" going on, there is "you don't understand what this particular English statement means". This isn't a math question, it's a language question. –  Robert Mastragostino Oct 1 '12 at 2:17

2 Answers 2

The current question seems to be this. Sorry, but this is "math language" with some prose sprinkled in.

Say I have 100 cupcakes selling for 1 dollar each. If I increase the number of cupcakes by 66%, then I'll have 166 cupcakes. $$\frac{\text{# of cupcakes}}{\text{price}}=\frac{166}{100}=1.66$$

If I have 100 cupcakes selling for 1 dollar each, and I instead decrease the price by 66%, then I'll be selling each one for 34 cents each, and we get $$\frac{\text{# of cupcakes}}{\text{price}}=\frac{100}{34}=2.94$$

So the issue seems to be this: Increasing the denominator by some percentage and decreasing the denominator by the same percentage don't give the same result. So what gives? Maybe this is just me, but I find that math helps me make this intuitive. Look at multiplying by a percentage like this: adding 66% is multiplying by 1.66=(1+.66), and subtracting 66% is like multiplying by .34=(1-.66). So if our percentage is $x$ (66% here), we find that

$$\frac{a(1+x)}{b}\neq\frac{a}{b(1-x)}$$

plug some numbers in to see that this is the same situation. Now $a$ and $b$ cancel, so we find that this is just saying that $$1+x\neq\frac1{1-x}$$

Now comes the real question: Why might people expect something different to happen? I think we need to look at what people think is happening here, where intuition works perfectly, and see why they're applying it somewhere where it doesn't belong.

Let's look at a different situation. The one people might think the above is: scaling. If we scale the numerator up by $k$, then we get

$$\frac{ka}b$$

If we scale the denominator down by $k$, we get

$$\frac a{\frac 1 k b}=\frac {ka} b$$

So here they are the same! This is what you might think the above situation was. But it clearly isn't. So what's the difference?

The difference is that you were talking about adding and subtracting percentages. So when you say you took off 66%, this is the reverse of adding 66%. But fractions don't work with addition in this way:

$$\frac{5+2}{3}\neq\frac{5}{3-2}$$

If you phrased it in terms of multiplication, everything would work. The reverse of scaling up is the reverse of scaling down. So let's repeat, but doing that instead.

Say I have 100 cupcakes selling for 1 dollar each. If I scale the number of cupcakes up by 1.66, then I'll have 166 cupcakes. $$\frac{\text{# of cupcakes}}{\text{price}}=\frac{166}{100}=1.66$$

If I have 100 cupcakes selling for 1 dollar each, and I instead scale the price down by 1.66, then I'll be selling each one for 34 cents each, and we get $$\frac{\text{# of cupcakes}}{\text{price}}=\frac{100}{\frac{100}{1.66}}=\frac{100}{60.24}=1.66$$


So the best I can say is that some people think that adding/subtracting percentages is the same thing as multiplying/dividing by them, which it isn't. A 66% decrease doesn't undo a 66% increase. This is because percentages are all about scaling a number, so talking about adding and subtracting them in the first place is really just horribly obscuring what's really going on. This I think is a language issue. Look at the differences here:

"Let's say I have 100 people who have 100 houses between them. If I double the number of people, then there will be 2 people to every house. If I halve the number of houses, there will be two people to every house."

"Let's say I have 100 people who have 100 houses between them. If I add 100% of people, then there will be 2 people to every house. If I subtract 100% of houses, there will be 2 people to every house."

Some careful reading would reveal that the second situation is wrong, and that the two are not the same. Doubling is the same thing as adding 100%, but halving is not the same thing as subtracting 100%. Talking about adding/subtracting percentages is just awkward and obscures what's really going on, which is scaling.

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Incidentally, for small percentages, "math language" makes it clearer that $\frac{1}{1-x}$ will be close to $1+x$. (They agree to first order near $x=0$.) For example, $0<\dfrac{1}{1-x}-(1+x) =\dfrac{x^2}{1-x}<2x^2$ when $0<x<0.5$. –  Jonas Meyer Oct 1 '12 at 5:13
    
no, only if you already think/read in math lang, not when you dont understand it, like what was just written completely bewilders me and everyone else, so foreign –  kittensatplay Oct 1 '12 at 5:28
    
will read and think carefully about this on non-sleeping time -- thanks -- forewarning the answerer though, that even if i understand everything bypassing the translation gap, i dont think it'll be satisfactory because the question "why" specifically requests a clear non-math answer/example, but we'll see, i may be able to translate the math into real-world, practical, helpful understanding –  kittensatplay Oct 1 '12 at 5:28
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If your'e working with numbers, you have to work with math. This is kind of like asking someone to speak French in Russian. There are some ideas that just don't translate properly (or at all) to other languages. Sometimes you just can't translate, and the only option is to gain some level of fluency/familiarity. –  Robert Mastragostino Oct 1 '12 at 5:35
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But you're not asking people to teach you math, you're asking people to give you English descriptions of math. If I explain that "scendere in pista" means "to take the floor" in Italian, then you aren't learning Italian (literally it's "to go down the trail") and you won't get anything useful out of it beyond the literal phrase I gave. If you want to really learn the concepts, you have to learn the language they're expressed in. Half the reason this is all so awkward and confusing is because you insist on seeing a 'translation' rather than the natural 'language' we're working in. –  Robert Mastragostino Oct 2 '12 at 3:29

Qualitative Approach

Consider it this way. Suppose you increase $A$ by $p$% giving $A^+$ and then decrease $A^+$ by $p$%. Because $A^+\gt A$, the $p$% decrease of $A^+$ will be greater than the $p$% increase of $A$, thus the end result will be less.

Likewise, suppose you decrease $A$ by $p$% giving $A^-$ and then increase $A^-$ by $p$%. Because $A^-\lt A$, the $p$% decrease of $A$ will be greater than the $p$% increase of $A^-$, thus the end result will again be less.

Quantitative Approach

$A(1-p)(1+p)=A(1-p^2)<A$.

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