[hanp]

In eyhung's excellent review of Jeff Rubens's book, he mentioned how you can compare two fractions, for example 4/13 and 5/16. The recommended method is to multiply 5*13 = 65 and 4*16 = 64, so 5/16 is bigger.

This is not how I usually do it. I naively subtract 4/13 from 5/16 to get (5-4)/(16-13) = 1/3. Clearly 1/3 (4/12) is larger than 4/13, so 5/16 is the larger one.

I have three questions:

1. Is this correct? (I know the answer to this one)

2. Do you think it is easier?

3. Does anybody else do it this way?

[jjbrr]

hi han

[TimG]

hanp, on Jan 14 2010, 04:02 PM, said:

This is not how I usually do it. I naively subtract 4/13 from 5/16 to get (5-4)/(16-13) = 1/3. Clearly 1/3 (4/12) is larger than 4/13, so 5/16 is the larger one.

You're saying that given a/b and c/d with a>c (and b>d),

if (a-c)/(b-d) > c/d, then a/b > c/d.

Have I correctly translated your process?

[hanp]

I think so.

I've never thought about it, this is just how I do it.

[jdonn]

I gave up. Maybe I'll have time at home. And it seems so simple too.

[bb79]

I think Han's method seems easier especially rationals close to 1... even if (a-c)/(b-d) and c/d hard to compare, you can iterate the process.

[gordontd]

I think eyhung's method is easier and more obvious than yours.

[hanp]

bb79, on Jan 14 2010, 05:09 PM, said:

you can iterate the process.

yup

[nigel_k]

A slight problem with Tim's formulation is that is doesn't require the converse to also be true, i.e. it should say 'if and only if' instead of just 'if'.

Anyway:
(a-c)/(b-d) > c/d
d(a-c) > c(b-d) (this is valid since factors are assumed to be > 0)
ad - cd > bc - cd
ad > bc
a/b > c/d

And this works in reverse as well.

[Fluffy]

my intuituive method on the given numbers would be to multiply both fractions by 3 and see that one is 1/13'th from 1, and the other is 1/16th, so the 5/16 must be bigger.

I learnt the cross product in school and I guess it is the easiest way to do it mathematically, I don't have my mind in the right state to try a counter example, but my bet is that han's method is wrong.

[PassedOut]

Fluffy, on Jan 14 2010, 05:29 PM, said:

I learnt the cross product in school and I guess it is the easiest way to do it mathematically, I don't have my mind in the right state to try a counter example, but my bet is that han's method is wrong.

I think this thread is an example of han's humor.

[matmat]

i just imagine cutting up cake into 15 or 13 pieces and deciding which of the two fractions of the cake i would rather have

(i'd actually rather not disclose what I do, then people might never ever want to think about playing with me again.)

[Elianna]

hanp, on Jan 14 2010, 01:02 PM, said:

In eyhung's excellent review of Jeff Rubens's book, he mentioned how you can compare two fractions, for example 4/13 and 5/16. The recommended method is to multiply 5*13 = 65 and 4*16 = 64, so 5/16 is bigger.

This is not how I usually do it. I naively subtract 4/13 from 5/16 to get (5-4)/(16-13) = 1/3. Clearly 1/3 (4/12) is larger than 4/13, so 5/16 is the larger one.

I have three questions:

1. Is this correct? (I know the answer to this one)

2. Do you think it is easier?

3. Does anybody else do it this way?

This clearly isn't something you'd want to teach people to do without any judgment.

Take for example the fractions 5/4 and 3/5. Whichever you decide is a/b and then do your subtraction you will get a negative number (which is clearly smaller than either fraction). This is clearly a contradiction (they can't be smaller than each other). You need to restrict that you subtract the one who's denominator is bigger by the one who's denominator is smaller.

Of course, you wouldn't use this method to check those, because you'd see that one was bigger than 1, and the other was less than 1.

Yours is of course true if b>d. Proof follows:

Assume (a-c)/(b-d)> c/d and (b-d)>0. Then, cross multiplying,

ad - cd > cb - cd and the cd's cancel so ad > cb, which is the same as eyhung's method and the method we were all taught in school.

[hanp]

Elianna, on Jan 14 2010, 08:52 PM, said:

This clearly isn't something you'd want to teach people to do without any judgment.

I wouldn't teach this to anybody, they might actually have to think!

[cherdanno]

I do it the same way actually (and never really thought about why I do it that way or why it is correct).
Seems quite natural to compute a determinant by using elementary row operations, now that I do think about it.

[kfay]

a/b :: c/d

(a+c)/(b+d) lies between the two

=> if (c-a)/(d-b ) < c/d then a/b < c/d

Yes?

[helene_t]

It took me some time to realize that what Han describes is the same as I do.

Easier? I dunno. May depend on the numbers.