I Prove that A:B is greater than C:D?

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The discussion focuses on proving that the ratio A:B is greater than C:D using known ratios M:N, A:B, and C:D. The argument is based on the inequalities A(M+1) > B(N+1) and D(N+1) > C(M+1), leading to the conclusions A/B > (N+1)/(M+1) and C/D < (N+1)/(M+1). These inequalities demonstrate that A:B is indeed greater than C:D, assuming all numbers are positive. The original poster expresses uncertainty about posting in the correct forum, but acknowledges the mathematical validity of the proof. The thread emphasizes the importance of clarity in mathematical discussions.
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From "Road to Reality" chapter 3 part two. If three ratios are known: M:N, A:B, C:D. A:B is greater than C:D if: A added to itself M times exceeds B added to itself N times and D added to itself N times exceeds C added to itself M times. Show that the ratio A:B is greater than C:D.
 
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I believe this should be posted in maths forum.
Based on the data,
A(M+1)>B(N+1) and D(N+1)>C(M+1)
So,
A/B>(N+1)/(M+1)...(1)
and
C/D<(N+1)/(M+1)...(2)
1 and 2 clearly prove A:B>C:D.
 
Thank you, it does belong to the mathematics category, but I'm unsure how to change it
 
Julia Coggins said:
but I'm unsure how to change it
You can't change it once you've posted it. A mentor will fix it.
 
Ah I see. Once again, thank you, I suppose for some reason I struggled with an otherwise simple question.
 
cnh1995 said:
I believe this should be posted in maths forum.
Based on the data,
A(M+1)>B(N+1) and D(N+1)>C(M+1)
So,
A/B>(N+1)/(M+1)...(1)
and
C/D<(N+1)/(M+1)...(2)
1 and 2 clearly prove A:B>C:D.
This is assuming all the numbers involved are positive.
 
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