why does multiplying a fraction e.g. enumerator/divisor with divisor/enumerator give 1?

Question

I just found out that if you want to get 1 with the fraction: $$\frac{5}{2}$$ Then you multiply it with: $$ \frac{2}{5} $$ Does anyone have a good way to think about this?

Answer 1

$$\frac{a}{b}\times\frac{b}{a}=\frac{a\times b}{b\times a}=\frac{ab}{ab}=1\quad\quad\quad a\neq 0\,,\, b\neq0$$

Answer 2

This is called The Inverse Property of Multiplication. Take $x$, then

$$x \cdot \frac 1 x = 1$$

In your case of $x=\frac 5 2$,

$$ \frac 5 2 \cdot \frac{1}{\frac 5 2} = \frac 5 2 \cdot \frac 2 5 = 1 $$

Answer 3

$x = a/b\iff bx = a.\,$ If there is a number $x'$ with $\,\color{#c00}{xx' = 1}$ then scaling this by $x'$ yields the

equation $\ b= b\color{#c00}{xx'} = ax',\,$ i.e $\,x' = b/a$

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More generally if $\,r\,$ is a root of a polynomial $f(x)$ then its reciprocal $\,1/r\,$ is a root of the reciprocal polynomial $\,\bar f(x) = x^n f(1/x),\ n = \deg f,\, $ obtained by reversing the coefficients of $f(x)$.

In the linear case we have $\ r = a/b\ $ is a root of $\,f(x)\, =\, \color{#c00}bx \color{#0a0}{- a}\,$

so $\,1/r\,$ is a root of $\,\bar f(x) = xf(1/x) = x(b/x-a) = \color{#0a0}{-a}x + \color{#c00}b,\,$ i.e. $\,1/r = b/a$

Note how the coeff list has been reversed $\,f = (\color{#c00}b,\color{#0a0}{-a})\, \mapsto\, \bar f = (\color{#0a0}{-a},\color{#c00}b).\,$ This is one way to explain why inverting a fraction swaps (reverses) the numerator and denominator.

Answer 4

Even before defining the rational numbers, one usually learns the 'prime factor cancellation' game, or how to divide without really trying.

For example, if you want apply Euclid's algorithm to $28$ and $16$, you can put $28$ 'on top' and $16$ on bottom, writing
$\frac{28}{16} = \frac{2^2\,7}{2^4} = \frac{7}{2^2}$
and then you can say that $16$ 'goes into' $28$ one and three-quarters times.

So, we have something that we can call the 'numerator/denominator' game. When you create the rational numbers, you hope that it would be helpful to keep this fractional notation. And indeed, it is very useful. If $n$ is a nonzero number, the multiplicative inverse $n^{-1}$ can be expressed as $\frac{1}{n}$ and you can continue playing the 'numerator/denominator' game in new ways.

The multiplication of fractional expressions is a blast (you can stick the prime factorizations together), but to add them you need a common denominator (no big deal).

For your question, just remember how much fun it is to cancel common terms in the numerator and denominator:

$\frac{5}{2}*\frac{2}{5} = \frac{5\;2}{2 \;5} = \frac{2 \; 5}{2 \; 5} \text { (cancel numerator factors /with denominator factors) } = 1$