My Math Forum Percentage Transformation

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 March 1st, 2016, 09:06 PM #1 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 Percentage Transformation I discovered ages ago that any percentage value has a counterpart. Let us say we have a number family, X, Y, and Z. Z is what percentage Y is of X. The thing with this number family, is that while knowing any two, the third is easy discern, X and Y can not be swapped. Well, it is quite often that I might know Z and one of the others, I might also want to know Z if Y and X were reversed. Basically, if Y is 125% of X, then X is what percentage of Y? If you know X or Y, you can find the other, flip them, then find Z again. But I have found the relationship between the two percentages. So now, there is a number family of four, X, Y, Z, and A. Z is the percentage Y is of X, and A is the percentage X is of Y. Z = 1- (A-1/A) No need for a reversal as you can swap A and Z and it still works (which makes sense of course). So what do you all think of this? Neat right?
 March 3rd, 2016, 12:06 AM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 Your formula doesn't make a lot of sense. Say $A = 80%$. Then $Z = 1 - 80 + \dfrac{1}{80} < 0$. How does that work?
 March 4th, 2016, 06:20 AM #3 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 How exactly did you derive that from my formula? What it should be, Z=1-(A-1/A) Z=1-(80-1/80) z=1-(79/80) z=1-(.9875) z=.0125 If you get something else, then a mistake was made somewhere.
 March 4th, 2016, 06:27 AM #4 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 Hmm, I see two mistakes, one mine one yours. I normally do programming math, so I forgot that division would be done before subtraction outside of programming. Your mistake was adding rather than subtracting, but could have been a typo.
March 4th, 2016, 06:00 PM   #5
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Quote:
 Originally Posted by MystMage I discovered ages ago that any percentage value has a counterpart. Let us say we have a number family, X, Y, and Z. Z is what percentage Y is of X. The thing with this number family, is that while knowing any two, the third is easy discern, X and Y can not be swapped.
That is true of any "non-commutative" binary operation.

Quote:
 Well, it is quite often that I might know Z and one of the others, I might also want to know Z if Y and X were reversed. Basically, if Y is 125% of X, then X is what percentage of Y?
You have swapped Y and Z here! If Y is 125% of X Then Y= 1.25X. To find solve for X, divide by 1.25= 5/4. X= Y/1.25= (4/5)Y= 0.80Y. X is 80% of X.

Quote:
 If you know X or Y, you can find the other, flip them, then find Z again. But I have found the relationship between the two percentages.
Yes, if Y is a% of X then then X is 1/a % of Y.

Quote:
 So now, there is a number family of four, X, Y, Z, and A. Z is the percentage Y is of X, and A is the percentage X is of Y. Z = 1- (A-1/A)
I'm not quite sure what you mean here. Are you saying then that Y is 50% of X then X is 1- (.5- 1/.5)= 1- .5+ 2= 3- .5= 2.5 or 250% of X? That is clearly not true. If Y is 50% of X then Y= 0.50X so X= (1/.50)Y= 2.00Y or X is 200% of Y.

Quote:
 No need for a reversal as you can swap A and Z and it still works (which makes sense of course). So what do you all think of this? Neat right?
If you mean what I suggested above, it is not correct. If you do not mean that what do you mean by "Z= 1- (A-1/A)"? If Y is 30% of X, what is your "A"? What is "Z"?

Last edited by Country Boy; March 4th, 2016 at 06:02 PM.

 March 4th, 2016, 07:01 PM #6 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 The error I refered to I guess wasn't clear, I didn't include a pair parenthesis, so try this, Z = 1- ((A-1)/A) Of course, you doing Y=ZX works, if you know two of those values. The equation I came up with is good if you don't know X or Y, just Z but want to see the percentage in the reverse direction.

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