# Are these fractions all equal?

Are the following expressions all equal to one another?

(2ab+c)/y

(2ab)/y + c/y

(2a)/y * (2b/y) + c/y


$$\dfrac{2ab+c}{y}=\dfrac{2ab}{y}+\dfrac{c}{y}=\dfrac{2a}{y}×\dfrac{2b}{y}+\dfrac{c}{y}$$

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Change 2b/y to b in the last one, then you're fine – gfes Jun 5 '11 at 12:55
$$\dfrac{2ab+c}{y}=\dfrac{2ab}{y}+\dfrac{c}{y}=\dfrac{2a}{y}b+\dfrac{c}{y}$$ – Américo Tavares Jun 5 '11 at 13:04
If you're having trouble seeing why the 3rd one is different, try putting $a=b=c=y=1$ into each expression. Note: expression, not equation; you haven't written any equations. – Gerry Myerson Jun 5 '11 at 13:05

By starting with the first, and applying the distributive law, we get the second: $$\frac{2ab+c}{y} = (2ab+c) \frac{1}{y} = 2ab \frac{1}{y} + c \frac{1}{y} = \frac{2ab}{y} + \frac{c}{y}$$

The last one is not equal to the other two, because $$\frac{2a}{y} \cdot \frac{2b}{y} + \frac{c}{y}= \frac{2a\cdot 2b}{y\cdot y} + \frac{c}{y} = \frac{4ab}{y^2} + \frac{c}{y},$$ and $\frac{4ab}{y^2} + \frac{c}{y}$ is (usually) not equal to $\frac{2ab}{y} + \frac{c}{y}$. Notice that we used the fact that $$\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}.$$ So the first two are equal, but not the third.

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Eivand's answer is complete - but I want to talk about a related idea. How does one know if something is true?

Often, proving that something is true is incredibly complicated. To prove something false, one needs only a counterexample. But the first step should always be to experiment with the idea. In this case, this probably means making a few fractions of the forms that you have described, and testing to see if they are equal.

After a few tests for the first 2 cases, you might convince yourself that they are true (this does not, of course, constitute a proof!). But the same heuristic approach would quickly destroy the third: suppose we let $a=1 \quad b=2 \quad c=3 \quad y = 5$.

Then $\frac{2a}{y} * \frac{2b}{y} + \frac{c}{y} = \frac{2}{5} * \frac{4}{5} + \frac{3}{5} = \frac{8}{25} = \frac{15}{25} = \frac{23}{25} < 1$.

On the other hand, $\frac{2ab+c}{y} = \frac{4+3}{5} > 1$. Almost any trial would yield a contradiction. I think that this style of patient problem-solving and analysis is under-emphasized and under-utilized, because it provides great insight on many questions.