Hi,

I'm not seeing how these two are equivalent: $\displaystyle -\frac{1}{x-1} \equiv \frac{1}{1-x}$.

Could somebody show/tell me, please?

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- Nov 24th 2010, 04:37 AMHellbentHow are these two equivalent
Hi,

I'm not seeing how these two are equivalent: $\displaystyle -\frac{1}{x-1} \equiv \frac{1}{1-x}$.

Could somebody show/tell me, please? - Nov 24th 2010, 04:45 AMemakarov
We have $\displaystyle a\cdot(-b)=(-a)\cdot b=-ab$. In particular, when $\displaystyle b = 1/c$, we have $\displaystyle \displaystyle\frac{a}{-c}=\frac{-a}{c}=-\frac{a}{c}$. In this case, $\displaystyle -(x-1)=1-x$.

Also, these two fractions are*equal*, not equivalent. - Nov 24th 2010, 04:52 AMTheCoffeeMachine
Or (more the same) just multiply both the numerator and the denominator by negative one:

$\displaystyle \displaystyle -\frac{1}{x-1} = \frac{-1}{x-1} = \frac{\left(-1\right)\left(-1\right)}{\left(-1\right)\left(x-1\right)} = \frac{1}{-x+1} = \frac{1}{1-x}$ - Nov 24th 2010, 05:56 AMWilmer
x = 3

-(1 / (3 - 1)) = -(1 / 2) = -1/2

1 / (1 - 3) = 1 / -2 = -1/2