Adding fractions with like denominators

Question

When we adding fractions why do not add denominators? For example, 2/15 + 3/15 = 5/15 not 5/30.

Answer 1

Because that is not how reality behaves. The number $2/15$ is meant to represent that you have divided something, let's say a pie, in $15$ equal pieces, and then from those you want to refer to only $2$ of those pieces.

So in this motivation, we have that the numerator (the number that goes in the upper part of the fraction) represents the number of pieces you have, while the denominator (the number in the lower part of the fraction) represents the size of each of those pieces.

Now, when you pick $2$ of those pieces and $3$ of those pieces, you end up having $5$ pieces, but the size of those pieces is the same (which is equal to the 15th part of the whole pie). So the numerator changes to reflect the change in the number of pieces but the denominator stays the same to reflect that the size of each piece has not changed.

Answer 2

Because $$\frac{2}{15} + \frac{3}{15} = 2\cdot \frac{1}{15} + 3\cdot \frac{1}{15} = \frac{1}{15}\cdot (2+3)=\frac{1}{15}\cdot 5 = \frac{5}{15}$$

Answer 3

Well, you could define it like that (there is just the problem of addition that way not being well-defined because of expanding of fractions: it would make $\frac12 + \frac11 \neq \frac12 + \frac22$, which is usually seen as bad). However, before you do, you should ask yourself two questions:

1. What is a fraction?
2. What is addition?

Most reasonable answers to those two questions, when taken together, will inevitably lead to the standard, established addition of fractions rather than your "easy" addition.

Answer 4

For all $a,b\in \mathbb{R}$ , $a \ne 0$, the equation $ax= b $ has a unique solution $x= a^{-1}b =: \dfrac{b}{a}$.

Your example: $a = 15,$ $ b=2,3.$

Rewriting:

$\dfrac{2}{15} = 2\dfrac{1}{15}$, with

$b=2, a^{-1} = \dfrac{1}{15}$.

$\dfrac{3}{15} = 3 \dfrac{1}{15}$, with

$b'=3, a^{-1} =\dfrac {1}{15}$.

$2\dfrac{1}{15} + 3\dfrac{1}{15} =$

(Distributive law)

$(2+3)\dfrac{1}{15} = 5 \dfrac{1}{15} =$

$\dfrac{5}{15}.$