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I really don't understand the expression.

$$\frac {1}{a}-\frac{1}{b}=\frac {b-a}{ab}$$

I generally have a hard time understanding non-intuitive things in math and this is one of them. Normally when I don't understand something I use an app , photomath, to help explain expressions/equations I don't understand however, I still need help with this expression.

I'm told that to get to $\frac {b-a}{ab}$ you need to expand the fraction to the least common denominator:

$$\frac {1}{a}-\frac{1}{b} \to \frac {\pmb b\times 1}{\pmb b a} - \frac {\pmb a \times 1}{\pmb a b} \to \frac {b}{ab} - \frac {a}{ab} \to \frac {b-a}{ab}$$

What I don't understand is this

$$ \frac {\pmb b \times 1}{\pmb b a} - \frac{\pmb a \times 1}{\pmb a b}$$

I don't understand how exactly the $a$ and $b$ seemingly 'appear' in the expression.

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    $\begingroup$ note that $ba=ab$ $\endgroup$ – J. W. Tanner Jul 7 at 1:25
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    $\begingroup$ Note: you aren't asked to find the $\textit {least}$ common denominator for $\frac 1a, \frac 1b$. $ab$ is $\textit {a}$ common denominator and that is all you need. $\endgroup$ – lulu Jul 7 at 1:30
  • $\begingroup$ If you are interested in the least common multiple see: en.wikipedia.org/wiki/Least_common_multiple $\endgroup$ – IntegrateThis Jul 7 at 1:31
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    $\begingroup$ multiplying $\frac{1}{a}$ by $\frac {b} {b}$ is the same as multiplying by $1$, similarly for multiplying $ \frac{1}{b}$ by $\frac{a}{a}$. $\endgroup$ – IntegrateThis Jul 7 at 1:32
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    $\begingroup$ Your algebra above is correctly executed. So, perhaps you can tell us which steps don't feel right to you? $\endgroup$ – Doug M Jul 7 at 1:38
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The golden rule for fractions is: you may multiply the numerator and the denominator by the same (non-zero) number without changing the value of the fraction. If, for example, you want to subtract $1/5$ from $1/3$, the golden rule enables use to do so: $$ \frac13-\frac15=\frac{5}{15}-\frac{3}{15}=\frac{2}{15}. $$ Or take $$ \frac{7}{9}=\frac{(\text{your favourite number})\cdot7}{(\text{your favourite number})\cdot9} $$ (in case your favourite number isn't zero). You may even write some strange things like $$ \frac{7}{9}=\frac{\frac{39}{101}\cdot7}{\frac{39}{101}\cdot9}, $$ but despite being useless it's not wrong anyway.

You can handle the general case in the same way: $$\frac1a=\frac{(\text{your favourite number})\cdot1}{(\text{your favourite number})\cdot a}. $$

Now let someone's favourite number be $b$.

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It is easier to grasp if you start with the RHS.

Note that $$\frac {b-a}{ab} = \frac {b}{ab} -\frac {a}{ab}=$$

$$ \frac {1}{a} -\frac {1}{b}$$

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$\frac{1}{a}-\frac{1}{b} = \frac{1}{a} * \frac{1}{1} - \frac{1}{b} * \frac{1}{1}$

= $\frac{1}{a} * \frac{b}{b} - \frac{1}{b} *\frac{a}{a}$ = $\frac{b} {ab} - \frac{a}{ab} = \frac{b-a}{ab}$

using $ab=ba$ and $\frac{x}{x} = 1$ for any $x \neq 0$

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a and b don't seemingly appear, they were there to begin with. First, you should note that in the original problem you have subtraction of two fractions (subtraction is not commutative order matters) and those fractions don't have a common denominator, which means you need to do some manipulation in order to combine the fractions into one.

Essentially what was done was that they multiplied (1/a) by this thing called a funny one. You can take something and multiply it by one and that doesn't change its value cause you can reduce it back to the original. So they multiplied it by b/b which by the way is 1. Same for the other fraction, they multiplied by a/a. Whatever you do to a denominator of a fraction you have to also do to it's numerator. When that's all done they were able to combine the fractions into one.

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Original Equation
$\frac{1}{a} - \frac{1}{b} \stackrel{?}{=} \frac{b - a}{ab}$

Multiply both sides by $ab$
$ab \cdot \big(\frac{1}{a} - \frac{1}{b}\big) \stackrel{?}{=} ab \cdot \big(\frac{b - a}{ab}\big)$

Distribute the $ab$
$\frac{ab}{a} - \frac{ab}{b} \stackrel{?}{=} \frac{ab \cdot (b - a)}{ab}$

Cancel out variables and both sides have identical expressions
$b - a \stackrel{\checkmark}{=} b - a$

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    $\begingroup$ The asker is clearly inexperienced in mathematics. Your idea is correct, but could you please show the steps? $\endgroup$ – K.defaoite Jul 7 at 4:01

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