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I came across this problem recently where

$A:b = C:A$ and $B= -9$ and $C = -4$

What is $A$ then?

I got $A^2 = 36$ $\Rightarrow$ $A= \pm 6$

However, as far as my knowledge goes a ratio is always positive

Therefore, we could easily eliminate $A=+6$ and the answer remains $A=-6$

Is my explanation correct, if not then why so? Can -ve Ratios exist?

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2 Answers 2

up vote 4 down vote accepted

Both $+6$ and $-6$ are correct answers. There is no rule that the ratio has to be positive.

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Your work is correct, but it's possible to have a negative ratio. Therefore, both $A = 6$ and $A = -6$ will satisfy the ratio.

It is true that a ratio between natural numbers will always be positive. That's because the natural numbers are the numbers $\{ 1, 2, 3, \ldots \}$. Since every natural number is positive, every ratio of natural numbers will be as well.

Since you're working with numbers like $-9$ and $-4$, however, it's clear that you're working with the set of integers, which are the numbers $\{ \ldots, -3, -2, -1, 0 , 1, 2, 3, \ldots\}$. Since some of the integers are negative and some are positive, we can definitely have a negative ratio between them.

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