# Multiplying whole number with fractions.

I'm looking at a solution to a math problem and there are obviously some rules regarding multiplication of fractions that I don't know.

Can someone make any sense of this?

$$s_n = 625 \cdot \frac{\left(\frac 1 5\right)^n - 1}{\frac 1 5 - 1} = 625 \cdot \frac{\left(\frac 1 5\right)^n - 1}{- \frac 4 5} = - \frac{3125}{4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = \frac{3125}{4} \cdot \left( 1 - \left(\frac 1 5\right)^n\right)$$

Can someone explain to me how you get from each step to another?

• Which part do you not understand? What do you know about multiplication and division of fractions? – Martin R Apr 20 '15 at 5:14
• I get step 1>step 2, that one's easy. It's how you get from step 2 to 3 and 3 to 4 that I don't understand. What is done with 625 and -4/5 to get -3125/4? – Julian Nikolay Krogh-Fredrikse Apr 20 '15 at 5:35
• Note that $a\cdot \dfrac{1}{\frac{2}{c}}=a\cdot \frac{c}{2}$. That's the principle being used there. – Sujaan Kunalan Apr 20 '15 at 5:39

First step: subtraction

Second step: $\frac{a}{\frac{b}{c}}=\frac{c}{b}a$

Third step: $(-c)(a-b)=(-1)c(a-b)=c(-1)(a-b)=c(-a-(-b))=c(b-a)$

The denominator $\frac{1}{5}-1 = \frac{1}{5}-\frac{5}{5}=\frac{1-5}{5}=\frac{-4}{5}$

Denominator's denominator is numerator, that makes $625 \cdot \frac{()}{-\frac{4}{5}}=-\frac{625 \times 5\times ()}{4} = -\frac{3125\times ()} {4}$

Lastly,$-\frac{3125}{4}=\frac{3125 \times \color{red}{-1}}{4}$ and when this $\color{red}{-1}$ taken inside the bracket $\left(\color{red}{-1} \times (\frac{1}{5})^n \color{red}{-1} \times -1\right)=\left( 1 - (\frac{1}{5})^n\right)$ because negative number multiplied by negative number becomes positive and positive number number multiplied by negative number becomes negative.

I'll try to lay out what happens in each step (equality).

Step 1. $\frac 1 5 - 1$ turns into $- \frac 4 5$. I assume you don't have a problem with this: it's just subtraction. Note that $1 = \frac 5 5$, so $\frac 1 5 - 1 = \frac 1 5 - \frac 5 5 = \frac{-4}{5} = - \frac 4 5$.

Step 2. The author does two things at once here: they multiply the fraction by $625$, and they carry out the division by $- \frac 4 5$. That might be confusing, so let's review what happens here. The first one is fraction multiplication. The big rule is: $$\frac a b \cdot \frac c d = \frac {ac}{bd}$$ Here's a good trick: $$a = \frac a 1$$ And this little rule follows from the trick and the big rule: $$b \cdot \frac 1 a = \frac b a = \frac 1 a \cdot b$$ The second topic, a little more complicated, concerns fraction division (the first equality here is just to be clear on notation): $$\frac {\frac a b} {\frac c d} = \frac{a/b}{c/d} = \frac a b \cdot \frac d c = \frac {ad}{bc}$$ One important thing about negation (this makes sense if you consider a negative sign to be the same as multiplying by $(-1)$): $$\frac {-a} b = - \frac a b = \frac a {-b}$$ So let's go through the equality slowly. We'll do the division part first.

$$625 \cdot \frac{\left(\frac 1 5\right)^n -1}{ - \frac 4 5} = 625 \cdot \frac{\frac{\left(\frac 1 5\right)^n -1}1}{ - \frac 4 5} = 625 \cdot \frac{\frac{\left(\frac 1 5\right)^n -1}1}{ \frac {4} {-5}} = 625 \cdot \frac{\left(\frac 1 5\right)^n -1} 1 \cdot {\frac {-5} 4}$$ Continuing, $$625 \cdot \frac{\left(\frac 1 5\right)^n -1} 1 \cdot {\frac {-5} 4} = 625 \cdot \frac{\left(\left(\frac 1 5\right)^n -1\right)\cdot(-5)} {1 \cdot 4} = 625 \cdot \frac{\left(\left(\frac 1 5\right)^n -1\right)\cdot(-5)} {4}$$

Then we rearrange: $$625 \cdot \frac{\left(\left(\frac 1 5\right)^n -1\right)\cdot(-5)} {4} = 625 \cdot \frac{(-5)} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = \frac{625 \cdot(-5)} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right)$$ Multiply... $$\frac{625 \cdot(-5)} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = \frac{-3125} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = - \frac{3125} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right)$$

Which completes that step.

Now the next step: we note that $$- \frac a b = \frac {-a} b = \frac{(-1) \cdot a}{b} = (-1) \cdot \frac a b$$

So we get: $$- \frac{3125} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = (-1) \cdot \frac{3125} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right)$$

And then we multiply the $(-1)$ with the right part of the equation. We can do this because: $$a \cdot b \cdot c = ab \cdot c = a \cdot bc = b \cdot ac = b \cdot a \cdot c$$ Et cetera. The point being is that when we multiply, the exact order in which we multiply the elements doesn't matter. (This is called commutativity.)

And now note that $(-1) \cdot (a-b) = b-a$. This makes sense because $(a-b) = (a + (-b))$, and $(-1) \cdot (a+(-b)) = ((-1) \cdot a) + ((-1) \cdot (-b) = ((-a) + (b)) = (b-a)$. Remember that $(-1) \cdot (-1) = 1$. So we get:

$$(-1) \cdot \frac{3125} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = \frac{3125} {4} \cdot (-1) \cdot \left(\left(\frac 1 5\right)^n -1\right)$$

And
$$\frac{3125} {4} \cdot (-1) \cdot \left(\left(\frac 1 5\right)^n -1\right) = \frac{3125} {4} \cdot (-1) \cdot \left(\left(\frac 1 5\right)^n + (-1)\right)$$ Now multiplying... $$\frac{3125} {4} \cdot (-1) \cdot \left(\left(\frac 1 5\right)^n + (-1)\right) = \frac{3125} {4} \cdot \left(\left( (-1) \cdot \left(\frac 1 5\right)^n\right) + ((-1) \cdot(-1))\right)$$ Finally, we're almost there... $$\frac{3125} {4} \cdot \left(\left( (-1) \cdot \left(\frac 1 5\right)^n\right) + ((-1) \cdot(-1))\right) = \frac{3125} {4} \cdot \left(-\left(\frac 1 5\right)^n + 1\right)$$ And of course: $$\frac{3125} {4} \cdot \left(-\left(\frac 1 5\right)^n + 1\right) = \frac{3125} {4} \cdot \left(1-\left(\frac 1 5\right)^n\right)$$

As desired.

• This ended up being a little longer than intended, but at least it was thorough. – Newb Apr 20 '15 at 5:34

Maybe you are confused about the multiplication of $\frac 1{5^n}$ with $-1$?

From first to second just substract $1$ from $\frac 15$.

Second part is to multiply $625$ with $\frac{-5}4$, which comes from $1/(-4/5)$ by dividing,

Finally we take $-1$ into the $(\frac 1 {5^n} - 1)$, and since every negative number will turn to positive and every positive number will turn to negative, each number will change its sign. In order to show that $n$ doesn't matter for our case,

for $n=1$: $-1 \cdot(1-1/5) = -4/5 = (1/5-1)$,

for $n=2$: $-1 \cdot (1-1/25) = -24/25 = (1/25-1)$,

As you can see, it doesn't matter whether n is odd or even we just do the simple multiplication.