I'm reading a book and there is a problem I need to clarify what the author wants me to do.

Show that $\frac{a}{b} = \frac{ak}{bk}$ follows from the law that $\frac{a}{b}=\frac{c}{d}$ if and only if $ad = bc$.

I've already proved the preceding law. My attempt to solve the aforementioned problem is as follows: $$ \begin{align} \frac{a}{b} &= ab^{-1} \\ &= ab^{-1} \cdot 1 \\ &= ab^{-1} \cdot kk^{-1}\\ &= ak \cdot b^{-1}k^{-1} \\ &= \frac{ak}{bk}. \end{align} $$

My question is is my approach is correct? I'm not sure because I didn't use the if and only if statement in my solution. Any hints?

  • $\begingroup$ No, your proof isn't correct. Reason: can you show the step where you used the law you were supposed to use? $\endgroup$
    – zipirovich
    Jan 8, 2017 at 5:37

1 Answer 1


I see my mistake and I do believe this is the correct solution

$$ \begin{align} \frac{a}{b} &= \frac{c}{d} \\ &= (cd^{-1}) \cdot (bb^{-1}) \cdot (kk^{-1}) \\ &= (cb) \cdot (d^{-1}b^{-1}) \cdot (kk^{-1}) \\ &= (ad) \cdot (d^{-1}b^{-1}) \cdot (kk^{-1}) \\ &= (ab^{-1}) \cdot (dd^{-1}) \cdot (kk^{-1}) \\ &= (ab^{-1}) \cdot (kk^{-1}) \\ &= (\frac{a}{b}) \cdot (\frac{k}{k}) \\ &= \frac{ak}{bk} \end{align} $$


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