Given positive integers $b, c, d, x, y$ such that: $\frac{a}{b}>\frac{x}{y}>\frac{c}{d}$ and $ad-bc=1$ prove that $x\ge a+c$ and $y\ge b+d$

Question

I was just trying to do the question above in the following way:

$ayd>bxd>bcy$ and $ad-bc=1$.

Hence we have that $ay>bx$ and $ad>bc$ and $xd>cy$.

This is where I got stuck. I couldn't continue from here. Could you please explain to me how to finish off my thought if this is possible, and if it isn't, please show another method for solving it whilst explaining the intuitive thought process behind each step?

The question as such has been answered here:How prove that $q \geq b+d$ for $ad-bc = 1$ and $\frac{a}{b} > \frac{p}{q} > \frac{c}{d}$? as pointed out in the comments. However I would like to know please if it is possible to finish it off using my method and if not being able to use my method fully then which part of my method could be assimilated into the solution?

Does this answer your question? How prove that $q \geq b+d$ for $ad-bc = 1$ and $\frac{a}{b} > \frac{p}{q} > \frac{c}{d}$? — Martin R, Oct 16 '20 at 11:35
no because I would like to know mainly if it is possible to finish off the question using my method — , Oct 16 '20 at 11:41
@Divide1918 yes they are. I have edited the problem statement — , Oct 16 '20 at 12:02
the people which are voting for the question to be closed, could you please explain to me why you are voting for it to be closed? I believe that my question is a lot different, as what I am asking is how to finish it off using my train of thought. Could you please explain to me why you believe that these questions are the same? — , Oct 16 '20 at 12:16
If this is what you are referring to then I just cross multiplied the denominators — , Oct 16 '20 at 12:18
@Michael: Note that this invalidates your further conclusions, such as $ac>bx$. — Martin R, Oct 16 '20 at 12:22
Yes, I realised right after my first edit. I have corrected it now. Thanks a lot @MartinR. Your help is much appreciated — , Oct 16 '20 at 12:24
@Michael: ... and now you can continue as in this answer https://math.stackexchange.com/a/897188/42969 to the referenced Q&A. Are you still convinced that your question is not a duplicate? — Martin R, Oct 16 '20 at 12:26
Yes @MartinR I still do believe that it is not a duplicate, as through this question, some other problems with my proof have also been resoled, problems which hopefully will help others as well — , Oct 16 '20 at 13:37

score 0 · Accepted Answer · answered Oct 16 '20 at 12:11

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Solution by WolfusA, we have $$0<\frac{a}{b}-\frac{x}{y}=\frac{ay-bx}{by} \implies ay-bx>0 \implies ay-bx \geqslant 1 \implies ady-bdx \geqslant d,$$ $$0<\frac{x}{y}-\frac{c}{d}=\frac{dx-cy}{dy} \implies dx-cy>0 \implies dx-cy \geqslant 1 \implies bdx-bcy \geqslant b. $$ Adding two last inequalities $b+d \leqslant y(ad-bc)=y.$

answered Oct 16 '20 at 12:11

Nguyenhuyen_AG

5,762

Who is “WolfusA”? – Martin R Oct 16 '20 at 12:15
Thanks a lot for your response :) – Oct 16 '20 at 12:24
This is the same solution as in https://math.stackexchange.com/a/897188/42969. – Martin R Oct 16 '20 at 13:40

Given positive integers $b, c, d, x, y$ such that: $\frac{a}{b}>\frac{x}{y}>\frac{c}{d}$ and $ad-bc=1$ prove that $x\ge a+c$ and $y\ge b+d$

1 Answers1