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If $$a,b,c,d,e,f$$ are positive integers and $$a/b<c/d<e/f$$.

Suppose $$af-be=-1$$ ,then prove that $$d>= b+f$$.

Note by Md Zuhair
1 month, 1 week ago

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$$\dfrac{a}{b} < \dfrac{c}{d} < \dfrac{e}{f} ;~ af - be = -1 \\ \dfrac{a}{b} < \dfrac{c}{d} \\ ad < bc \\ \boxed{ad + x = bc} \\ \dfrac{c}{d} < \dfrac{e}{f} \\ cf < de \\ \boxed{cf + y = de\\ cf = de - y} \\ \boxed{af - be = -1 \\ af = be-1}\\ ad + x = bc \\ adf + xf = bcf \\ d(af) + xf = b(cf) \\ d(be-1) + xf = b(de-y) \\ dbe - d + xf = dbe - by \\ xf - d = -by \\ d = xf + by \\ \boxed{d \ge b + f}$$ · 1 month, 1 week ago

Great! The interesting fact is that $$d = xf + by$$, which is extremely restrictive. Staff · 1 month, 1 week ago

Yeah, restricts $$x, y \in R$$ · 1 month, 1 week ago

You are going to top TN Region RMO This time @Vicky Vignesh · 1 month, 1 week ago

What have you tried? What do you know? Staff · 1 month, 1 week ago

I did'nt understood your question? · 1 month, 1 week ago

How have you tried to solve this problem?

For example, do you know if $$d > b$$ or $$d > f$$? Staff · 1 month, 1 week ago

I could'nt do it. · 1 month, 1 week ago

@Vicky Vignesh .. Could you? · 1 month, 1 week ago

What have you tried?

If you've tried nothing, then go try something. What approaches can you think of? What comparisons can we make? Staff · 1 month, 1 week ago

I started by assuming d<b+f. Then I did lot of manipulations and after a while a wrong result came up which means this is a contradiction. So d must be >=b+f. Is this correct? · 1 month, 1 week ago

Assuming that the steps are correct, then that sounds like a possible proof approach. You should write up the solution and others can comment on it. Staff · 1 month, 1 week ago