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Is it true that, for all positive $A$, $B$, $C$, $D$, if $A-B>C-D$ then $AD>BC$ ?

I'm just wondering about this, and I tested it with a bunch of integers and haven't found a counterexample yet. Is there a way to know if it's true?

If it is true, is it true for all positive reals, or just natural numbers?

(Here's how I'm asking myself in my head: "So you have two quantities, and each of them has two factors. A factor of the first quantity is larger than a factor of the second quantity by more than the amount by which the other factor of the first is smaller than that of the second. Is the first quantity always greater?")

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  • $\begingroup$ Gah...I'm sorry, everyone. This was silly. $\endgroup$ – Annick Oct 9 '12 at 21:41
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$7-3>4-1$ but is $7\cdot 1>4\cdot3$?

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  • $\begingroup$ If you wanted all positive... $\endgroup$ – Dennis Gulko Oct 9 '12 at 21:39
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What about $0 - 1 > 1 - 4$? Certainly $0\cdot 1 \not > 1 \cdot 4$.

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In the title of the article the desired conclusion is $AB>CD$, while in the body the first line refers to the conclusion $AD>BC$. What follows assumes it's the latter conclusion being asked about. [The discussion about motivation seems to me to support that this is the desired conclusion.]

What I found is that, if one assumes not only that $A-B>C-D$, but also that the differences of the squares go in the opposite direction, i.e. $A^2-B^2<C^2-D^2$, then one can actually conclude that $AD>BC$.

For $A,B,C,D$ positive, $A-B>C-D$ iff $A+D>B+C$ iff $(A+D)^2>(B+C)^2.$ The last statement is equivalent to

[*] $A^2+D^2-B^2-C^2+2(AD-BC)>0$.

So if we assume not just $A-B>C-D$ but also the $opposite$ for the squares, namely $A^2-B^2<C^2-D^2$, we get to $A^2+D^2-B^2-C^2<0$, and hence using [*] we get to $2(AD-BC)>-(A^2+D^2-B^2-C^2)>0$ and finally to $AD>BC$. So with the extra assumption about the squares, we can get the relation between the products.

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