Complex fraction

my book uses a technique to write fractions in a way that I am not familiar with.

How does $$\frac{\frac{1}{243}-1}{-\frac{2}{3}}$$ equate to $$\frac{3\left(1-\frac{1}{243}\right)}{2}$$ Thanks in advance.

Division means multiplication of the reciprocal, so $$(\frac1{243}-1)\div (-\frac23)\implies(\frac1{243}-1)\times(-\frac32).$$

As well, a binomial involving subtraction can be rewritten by factoring out a negative: $$(a-b)=-(b-a).$$

Combining the two ideas, we have \begin{align} (\frac1{243}-1)\times(-\frac32)&=(1-\frac1{243})\times(\frac32)\\ &=\frac32\times(1-\frac1{243})\\ &=\frac{3(1-\frac1{243})}{2} \end{align}

• Exactly the very last step I did not understand. I would assume your last step is a result of $$\frac{3}{2}*\frac{\left(1-\frac{1}{243}\right)}{1}$$ , am I right? – Math Noob Nov 6 '19 at 19:47
• Yes, precisely. Any expression $x$ can be written rationally as $\frac{x}1$. Then you can proceed to multiply the numerators and denominators. – Andrew Chin Nov 6 '19 at 19:48
• Although, I would argue that it would be simpler just to evaluate $1-\frac1{243}=\frac{242}{243}$ and take it from there. I'm not sure why your book does not do this. – Andrew Chin Nov 6 '19 at 19:50

The numerator and denominator of the fraction were multiplied by $$-3$$. That clears the complex fraction in the denominator and inverts the binomial in the numerator.

You multiply numerator and denominator both by $$-3$$ ("expand the fraction by a factor of $$-3$$"). This operation is well-known not to change the value of a fraction.