A question from 1964!

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ScholMaths

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This is a question from a 1964 scholarship examination paper I'm struggling to complete:

The positive integers a,b,c,d,e,f are such that:

a/b > c/d > e/f and af-be=1

Prove that 0 < (cf-de)/(df) < 1/(bf) and hence show that d>b and similarly that d>f

I can prove the first part as a/b - e/f is greater than c/d - e/f. Work through the algebra and use af-be=1 to get the result.

I can't then show that d>b and d>f.

Any help/hints appreciated.
 
Hi,

Let me show you my work:

a/b>c/d>e/f

(af)/b>(cf)/d>e

(af)>(bcf)/d>(be)

since (af)=(be)


(be)>(bcf)/d>(be)

e>(bcf)/d>e

How can this be true for b, c, f and d?

To me there is something wrong with the question?

Hope this helps.
 
So you know

1/(bf) > (cf - de)/(df) > 0

Multiply through by df

d/b > cf - de > 0

Since c,f,d,e are positive integers and cf - de > 0, cf - de is an integer greater than 0 (the linear combination of integers will again be an integer). So,

d/b > cf - de >= 1 (Since 1 is the smallest integer greater than 0).

Then d/b > 1 and d > b.
 
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