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## Monday, January 28, 2008

### Artful Arithmetic

Professor Egghead had a student who was not very good with fractions and thought she had stumbled upon a quick way of discovering which of two fractions was the larger.

When she was asked to find the larger between 2/5 and 3/7 she simply subtracted the numerator from the denominator in each fraction, replacing them with 2/3 (2/(5-2)) and 3/4 (3/(7-3)) respectively, which she then replaced with 2/1 and 3/1, using the same method, and concluded that the first, 2/5, was the smaller.

Professor Egghead was impressed with her method. Was her method valid or was it complete nonsense and her correct answer only a lucky coincidence.

Loralie said...

3/7 5/8
3/4 5/3
3/1 5/-2

That's the one I just made up and tried, and it doesn't see like her way quite worked. If you stopped at the second line it did, though, and I've tried some other fractions where it worked. I THINK it's just a coincidence though.

C said...

If we consider subtractions in denominators, we have 2/5 and 3/7,
becomes 3/5 and 4/7,
then becomes back 2/5 and 3/7 (infinite work, stop)
However, we can see that 2/5 < 3/7, but 3/5 > 4/7 (they are in inverse proportion)
This way is infinite and inverse because the differences are replaced with the smaller factors - the substituted numerators.

Contradiction with this, when differences are replaced by larger factors - the denominators, new fractions should be in direct proportion with old fractions.

It means the method of that student is actually valid.
Although the method can not be proved in mathematical way with numbers, it can be logically proved based on mathematical characteristics. Anonymous said...

It is a total coincidence:

the student has gone about solving it in 2 steps -
1) keep numerator the same but change denominator to (denom-num)
2) keep numerator the same but change the denominator to 1

As you can see from that, step 1 is completely redundant so she is basically comparing the sizes of the numerator.
Obviously some of the time this will give a correct answer but not for any valid reason.
Try it for yourself using 7/12 and 5/7 for instance Anonymous said...

it,s a total coincidence.
Try it with fractions greter than 1....

Erik said...

This method of finding the larger fraction a/b ? c/d (subtracting the value of the numerator from the denominator on both sides of the inequality) is valid as long as the following conditions are true: a, b, c, d <> 0 and (b-a), (d-c) <> 0.

Let's say that the fraction on the left side is the SMALLER of the two fractions being compared:

Step 1. a/b < c/d

If we apply the multiplicative inverse to both sides (take the reciprical of each fraction, or flip each fraction upside down), then the inequality changes (a/b becomes b/a and c/d becomes d/c and the left side is now LARGER than the right):

Step 2. b/a > d/c

Note here that if the fractions were equal to begin with, they will remain equal even after applying the multiplicative inverse.

We can now subtract 1 from both sides of the inequality:

Step 3a. b/a - 1 > d/c - 1

or

Step 3b. b/a - a/a > d/c - c/c

This is the same as:

Step 3c. (b-a)/a > (d-c)/c

Applying the multiplicative inverse to both sides again leaves us with:

Step 4. a/(b-a) < c/(d-c)

This means that Professor Egghead's student was right. Let's try it with the examples used in the comments above.

In loralie's example:

3/7 ? 5/8

Subtracting the numerator from the denominator leaves:

3/4 ? 5/3

At this point we can stop because 3/4 is less than 1 and 5/3 is greater than 1. It's easy to see that 5/3 is greater than 3/4, therefor 5/8 is greater than 3/7:

3/7 < 5/8

Anonymous wants to try the fractions 7/12 and 5/7:

7/12 ? 5/7

Subtract the numerator from the denominator on both sides:

7/5 ? 5/2

At this point we can see that 7/5 is larger than 1 but less than 2 and 5/2 is larger than 2. Therefor we can conclude that 5/2, and by association 5/7, is the larger fraction:

7/12 < 5/7

Anonymous Two wants to try fractions that are larger than 1. Basically we are already doing that in Step 2 of the proof. We only need to remember to reverse the inequality sign after reaching our conclusion in order to compensate for the implied multiplicative inverse.

Compare 12/5 with 7/3:

First take the inverse of both fractions to make them both less than 1 then continue as before:

5/12 ? 3/7 =>

5/7 ? 3/4 =>

5/2 ? 3/1

Here we can see that 5/2 is larger than 2 but less than 3 and 3/1 is equal to 3, therefor 3/1 is larger, HOWEVER, remember that we took the inverse of each original fraction so now we have to reverse our inequality sign when going back to the original fractions:

12/5 > 7/3 Josh said...

In order for this to work the way the student has done it, both denominators must be odd and the numerators must be equal to (Respective Denominator/n)+1 where n can be any number but must be the same for both fractions.

If this is the case than yes the method works, but how often is that going to happen? JDK said...

It works... WTF? Here is the proof. I did not go through and find the values that make it fall apart but ill note the step.
First we need to see that the student's theorem is:
if a/b > c/d
then a/(b-an) > c/(d-cn)
where n = any int

so a/b > c/d
= ad/bd > bc/bd
= ad > bc
= ad - acn > bc - acn
= a(d-cn) > c(b-an)
= [a(d-cn)]/[(b-an)(d-cn)]
> [c(b-an)]/[(d-cn)(b-an)]
***Dividing by 0?***
= a/(b-an) > c/(d-cn)
QED Apple prodam iphone said... Anonymous said...

COOL!!!!

Abdulelah Fallatah said...

Let us swap the nom with the denom in 3/7.

2/5 < 7/3
2/3 < 7/-4 (FALSE)