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$a > b$ $(a - b)(b - c)(c - a) > 0$ Which is bigger, $a$ or $c$?

4 Answers 4

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$a-b$ is positive, so $b-c$ and $c-a$ have the same sign. Then $$ac$.

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    ummm, We are given $a>b$, so the first option is false, so $c\lt a$. – 2017-01-22
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    True. Just a silly mistake. Thanks. – 2017-01-22
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Let us assume $c>a>b$. Then, we have $(a-b)(b-c)(c-a)<0$ because of $a-b>0,b-c<0,c-a>0$.

Therefore we can rule out $c>a$.

$a=c$ would imply $(a-b)(b-c)(c-a)=0$, so is impossible as well.

Therefore we can conclude $c

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    If we have $b0$ , $b-c<0$ , $c-a<0$, so in this case, both conditions are met. – 2017-01-22
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Let $a\lt c$

It follows that $b\gt c$ so that the initial condition

$(a-b)(b-c)(c-a)\gt 0$ holds true. ( If two of the terms within the parentheses are positive, the other one must be as well )

But $a\gt b$. A contradiction. Thus $a\gt c$.

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    Thanks for taking your time! You say "It follows that $b > c$" but it should be reversed if $a < c$ (from your first assumption it should be like $c < a < b$). – 2017-01-22
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    @My pleasure! Well, methinks it shouldn't since that "so that" is present in the sentence linking the argument.. – 2017-01-22
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    @Estonian I was curious about that too. It *does* follow that $b > c$ by considering signs (and it's worth mentioning *something* to that effect). So in reality, the contradiction has already shown itself, as this violates transitivity of $<$. – 2017-01-22
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    @pjs36 I dare say we are caught in a "nominalistic" argument. I suppose I could have written: It follows that $b\gt c$ so that the initial condition $(a-b)(b-c)(c-a)\gt 0$ is true.. – 2017-01-22
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    Yeah, I think you're right. I was expecting the justification *before* the conclusion; so I lost my train of thought a bit and couldn't process how it all came together. – 2017-01-22
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Remember $k < j; x > 0$ then $kx < jx$ and $k/x < j/x$ and if $k < j; y < 0$ then $ky > jy$ and $k/y > j/y$.

Or in imformal terms "multiplying or dividing by a positive keeps the less than/greater than sign; multiplying or dividing by a negative flips it".

$a > b$ so $a - b>0$.

So if $(a-b)(b -c)(c-a) > 0$ then dividing be $a-b$ will "keep the sign"

$(b -c)(c-a) > 0$. So either $(b-c)$ and $(c-a)$ are both positive; or they are both negative.

If they are both positive then $b > c$ and $c > a$. SO $b > c > a$ and $b > a$. But we know that is not true.

If the are both negative then $b b$ which is consistant with what we already know

So: $a > b; c>a;$ and $c>b$ so

$(a-b) > 0$

$(a-b)(b-c) < 0$ and

$(a-b)(b-c)(c-a) > 0$.