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Suppose $a,b,c,d \in \mathbb{R}^+$ are such that $$ a \leq b \\ c \leq d \\ ac = bd $$ then is the following true? $$ a= b \\ c = d $$

Attempt:

Suppose not true, i.e. $a < b$ or $c < d$, then we have $ac < bd$ thus contradicting the fact that $ac=bd$. Thus $a= b$ and $c = d$.

So, is my argument correct? Sorry for posting this if this is a trivial fact.

Random-generator
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    It's fine for me. – Bernard Jul 20 '17 at 19:13
  • @Bernard Thanks for the confirmation. Should I leave this post as it is or delete it? Asking this since this may be trivial for some and they might downvote the same. What do you suggest? – Random-generator Jul 20 '17 at 19:15
  • I think you need to consider that a and c need not be positive. -1 < 1. -2 < 2 and (-1)(-2) = 1x2 so no. But if you have $0 \le a \le b$ and $0 \le c \le d$ then you would have $a < b \implies ac < bc < bd$ so $ac =bd $ would have your result. – fleablood Jul 20 '17 at 19:16
  • @fleablood Thank you. But notice that $\mathbb{R}^+$ I had in the question details. – Random-generator Jul 20 '17 at 19:17
  • Ah... then you are golden. Remember $x < y; b > 0 \implies bx < by$ and so $a < b \implies ac < bc \le bd$ etc.... – fleablood Jul 20 '17 at 19:20
  • @Lohith-kumar I'd rather write that as $,a,b,c,d \gt 0,$ to make it unambiguously clear that all numbers are assumed to be strictly positive. The notation $,\mathbb{R}^+,$ can sometimes be used to refer to the non-negative reals, including $,0,$ (in which case the assertion fails). – dxiv Jul 20 '17 at 19:23
  • I think it's a example of a proof by contrapositive. It might be helpful to those who confuse contrapositive and proof by contradiction. – Bernard Jul 20 '17 at 19:25
  • @dxiv: positive real numbers should be denoted $\mathbf R^{+*}$. $\mathbf R^+$ is used for the non-negative. – Bernard Jul 20 '17 at 19:26
  • I wouldn't say it's "trivial" although it is easy. And even with stating $a,b,c,d \in \mathbb R^+$ it is still important to pin down the precise conditions. As $a = b = 0$ yeilds no conclusion about $c$ and $d$. – fleablood Jul 20 '17 at 19:27
  • @Bernard Good point. – Random-generator Jul 20 '17 at 19:28
  • @dxiv Oh, thanks for pointing it out. – Random-generator Jul 20 '17 at 19:28
  • @Bernard There are quite a few different opinions and notations here for example. It's always safest to define the terms whenever there is even a remote chance of ambiguity. – dxiv Jul 20 '17 at 19:29
  • i think in this case, we need to specify conditions. There are a finite number of conditions between whether a,b,c,d can be pos, neg, or zero, but the spirit of this result applies to strictly positive with specific exceptions figured out directly. – fleablood Jul 20 '17 at 19:49

3 Answers3

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Okay, FINAL final answer:

If $0 \le a \le b$ and $0 \le c \le d$ then

$ac \le bc$ with equality holding only if $a = b$ or $c = 0$.

And $bc \le bd$ with equality holding only if $c = d$ or $b= 0$.

So $ac \le bd$ with equality holding only if one of the four occur:

1) $a = b$ and $c = d$

2) $a =b$ and $b = 0$ so $a = b =0$.

3) $c = 0$ and $c= d$ so $c = d = 0$.

4) $c =0$ and $b= 0$ and $0 \le a \le b = 0$ so $a = b = c = 0$

So if $ac = bd$ then either i) $a=b$ and $c=d$ or ii) $a=b=0$ or $c=d=0$.

And if $ac = bd \ne 0$ then $a=b$ and $c=d$.

Otherwise $ac < bd$.

fleablood
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Alternatively: let $b=ax, x\ge 1$ and $d=cy, y\ge 1$. Then: $$ac=bd \Rightarrow ac=axcy \Rightarrow xy=1, x\ge 1, y\ge 1 \Rightarrow x=1, y=1.$$

farruhota
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$$0=bd-ac=bd-bc+bc-ac=b(d-c)+c(b-a)\geq0.$$

Hence, for positive variables we have $a=b$ and $c=d$.

If our variables are non-negatives we need $b(d-c)=0$ and $c(b-a)=0$, which gives some cases.

Michael Rozenberg
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