Math Is Fun Forum

lunatic

Guest

fractions

Hiya,
Need some help in proving something..

if we consider fractions like xy/yx

for example 43/34 it is equal to

43/34 = 473/374 = 4773/3774 = 47773/37774 etc

Basically the number in between is the addition of the 2 numbers 4 and 3 whcih is 7.

how can i prove that for any number of 7's inbetween the, value of the fraction remains the same..

any help??

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Re: fractions

Quite interesting.

I don't know if this will be any help to you, but you're just multiplying the top and bottom by 11, 111, 1111, 11111 etc.

Last edited by Toast (2007-01-03 12:01:10)

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: fractions

Quite interesting.

I don't know if this will be any help to you, but you're just multiplying the top and bottom by 11, 111, 1111, 11111 etc.

Excellent! Good obsrvation, toast!
The answer is simple: just:
477...773 = 11..11*43 (k 7's and k+1 ones)
377...774 = 11..11*34 (k 7's and k+1 ones)

But it's interesting.

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IPBLE:  Increasing Performance By Lowering Expectations.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: fractions

477...773 = 11..11*43 (k 7's and k+1 ones)
377...774 = 11..11*34 (k 7's and k+1 ones)

Doesn't that need to be proved?

I was too, very puzzled by this at first.  I attempted to do it with modulus, with no avail.  After about 5 minutes, I realized what it means for two rational numbers to be equal.  Simply put, if we have two rational numbers:

a/b and c/d

a/b = c/d iff ad = bc

43*374 = 473*34

Or in general:

4(7 k times)3 * 3(7 k+1 times)4 = 4(7 k+1 times)3 * 3(7 k times)4

The rest of the proof is simply the base 10 expansion of a number, then multiplying terms.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."