# Is $15/52$ equal to $17/59$?

Is $\frac{15}{52} = \frac{17}{59}$? I typed it into the calculator and found: $$\frac{15}{52} = 0.2884615$$ $$\frac{17}{59} = 0.2881356$$ So I thought they were different. But then my friend said that the small difference was due to the roundoff error of the calculator, and they are in fact equal. Is my friend right?

– user228113
May 21, 2015 at 1:12
• This reminds me. May 21, 2015 at 1:13
• In general, if you have two irreducible fractions (which both $\frac{15}{32}$ and $\frac{17}{59}$ are), they are only equal if their numerators and denominators are equal. May 21, 2015 at 1:40

The Theorem of Cross-products says that if $\frac{a}{b} = \frac{c}{d}$, then $ad = bc$. This is easily verified by multiplying both sides of $\frac{a}{b} = \frac{c}{d}$ by $bd$.

Applying this to your problem, $\frac{15}{52}$ would only equal $\frac{17}{59}$ if $15\times 59 = 17\times 52$. This turns out to be NOT true, since: $$15\times 59 = 885$$ $$17\times 52 = 884$$

Although $885$ and $884$ are close (which actually turns out to be the case because $\frac{15}{52}$ and $\frac{17}{59}$ are so close to each other), they are NOT equal.

Your friend is not right, and the calculator does give the correct values (there is a difference of $\frac{1}{3068} = 0.000326$, found by subtracting the fractions).

• Using prime numbers decomposition, you don't even need to compute $15\times 59$ and $17\times 52$ to check they are different. The latter is a multiple of $17$, contrary to the former. May 21, 2015 at 1:48
• That is true. I could have explained it that way. However, I wanted to show the OP that the fact that the two products $884$ and $885$ are so close means that the original two fractions were close.
– Ant
May 21, 2015 at 1:51
• Since someone learning might not immediately recognize 17 as a prime, it might be easier to argue that 52 is even and 59 is odd. May 21, 2015 at 3:34

Here is a picture to help with your intuition. (It's not rigorous like the other answers. For instance, you have to trust me that the boxes in each row are all the same width.) And here's the source code used to produce the graphics.

\documentclass{minimal}
\usepackage{tikz}
\usetikzlibrary{calc}

\def\boxes{
\begin{scope}[shift={(0, 2)}]
\draw[gray] (0, 0) -- (0, 3);
\draw[gray] (59, 0) -- (59, 3);
\draw[fill=green] (0, 0) rectangle (15*59/52, 2);
\draw (0, 0) rectangle (59, 2);
\foreach \x in {1, ..., 51} {
\draw (\x*59/52, 0) -- (\x*59/52, 2);
}
\end{scope}
\draw[gray] (0, 0) -- (0, -1);
\draw[gray] (59, 0) -- (59, -1);
\draw[fill=blue!80] (0, 0) rectangle (17, 2);
\draw (0, 0) rectangle (59, 2);
\foreach \x in {1, ..., 58} {
\draw (\x, 0) -- (\x, 2);
}
}

\begin{document}
\begin{tikzpicture}[scale=0.25]
\boxes
\node[below] at (0, -1) {$0$};
\node[below] at (59, -1) {$1$};
\draw[red] (17, 2) circle (2.5cm);
\begin{scope}[scale=6.0, shift={(-12, -5)}]
\clip (17, 2) circle (2.51cm);
\draw[red] (17, 2) circle (2.5cm);
\boxes
\foreach \x in {14, ..., 17} {
\node at ({(\x - 1/2)*59/52}, 3) {$\frac{\x}{52}$};
}
\foreach \x in {16, ..., 19} {
\node at ({\x - 1/2}, 1) {$\frac{\x}{59}$};
}
\end{scope}
\end{tikzpicture}
\end{document}

• What program did you use to create this figure? It is a very nice visual representation.
– Ant
May 21, 2015 at 21:31
• I used the TikZ package in $\LaTeX$. I can edit the answer to include the source code. May 25, 2015 at 1:14

Just in case you're genuinely curious, one easy technique to verify whether it's right or not is to cross-multiply. The equality $\frac{15}{52} = \frac{17}{59}$ holds if and only if the equality $15 \cdot 59 = 17 \cdot 52$ holds.

We notice that $15 \cdot 59 = 885$ while $17 \cdot 52 = 884$, so no, the two fractions are not equal.

any time you want to check whether two fractions $\color{red}{\large{\frac{a}{b}}}$ and $\color{blue}{\large{\frac{c}{d}}}$ are the same or not. You just need to check if $\color{red}{a}\color{blue}{d} = \color{red}{b}\color{blue}{c}$

if $$\color{red}{a}\color{blue}{d} = \color{red}{b}\color{blue}{c}$$ then $$\color{red}{\large{\frac{a}{b}}} = \color{blue}{\large{\frac{c}{d}}}$$

else $$\color{red}{\large{\frac{a}{b}}} \neq \color{blue}{\large{\frac{c}{d}}}$$

• haha thanks @ken May 21, 2015 at 2:33

Your friend is messing with you - one way to see they have to be different is to note that 17 and 59 are both prime. EDIT: this means that the fraction ${17\over 59}$ is in simplest terms, and can't be equal to any fraction with smaller numerator and denominator - such as ${15\over 52}$. Note that this is a different approach from cross-multiplying: which is 'better' all depends on whether you want to look at prime factors, or do multiplication.

(Frankly, I'd be amazed if there were a calculator in existence these days with such bad roundoff errors. :P)

If a fraction $\frac{p}{q}$ has been expressed in its lowest terms, i.e. if the numerator $p$ and denominator $q$ have no common factors (which is true for both of the fractions in this example, $\frac{15}{52}$ and $\frac{17}{59}$) then the values of the numerator $p$ and the denominator $q$ are uniquely determined by the value of the fraction.

That is to say, if two fractions $\frac{p}{q}$ and $\frac{r}{s}$ are both expressed in their lowest terms, and if $\frac{p}{q} = \frac{r}{s}$, then $p = r$ and $q = s$.

Because in this case $p \ne r$ (and $q \ne s$), it cannot be true that $\frac{p}{q} = \frac{r}{s}.$

When the numbers involved are fairly small (as they are in this case), it is quite feasible to check mentally whether both of the fractions are in their lowest terms, or not; and if they are not, then to cancel out any common factors, reducing both the fractions to their lowest terms, so you can tell simply by inspection whether they are equal or not.

But with larger numerators or denominators, it is probably simpler and quicker to multiply out, as has been explained in other answers.