Question

How would you convert the repeating, non terminating decimal into a fraction? Explain the process as you solve the problem 0.1515

Answer 1

@Hero

Answer 2

.
First, count how many are not repeating: 0
Subtract the number from the whole decimal: 0.15-0=0.15
Now, put a 9 for every number repeating
The answer is 15/99
Reduce: 5/33

Answer 3

this good?

Answer 4

Let

x = 0.1515...
100x = 15.1515...

100x - x = 15
99x = 15
x = 15/99
x = 5/33

Answer 5

ok so 100*0.1515

Answer 6

yes

Answer 7

But what were you going to say @amistre64 ?

Answer 8

i was thinking of 15/100+15/10000+15/1000000+... and decided it was just not worth the effort to try to go on :)

Answer 9

@amistre64, you never cease to humor me. I don't know whether you're just 'funnin' or not.

Answer 10

a geometric sum was all i was considering ;)

Answer 11

I just simply did was I was taught to do. I wasn't aware of any other methods to solve.

Answer 12

there is a pattern:\[\frac{15}{100}+\frac{15}{100^2}+\frac{15}{100^3}+\frac{15}{100^4}+...\]
\[15\left[(\frac{1}{100})+(\frac{1}{100})^2+(\frac{1}{100})^3+(\frac{1}{100})^4+...\right]\]
\[\frac{15}{100}\left[1+(\frac{1}{100})+(\frac{1}{100})^2+(\frac{1}{100})^3+(\frac{1}{100})^4+...\right]\]
this gives us a geometric series in 1/100
\[\frac{15}{100}\lim_{n\to \infty}\frac{1-.01^n}{1-.01}\]
\[\frac{15}{100}\frac{1}{1-.01}\]
\[\frac{15}{100}\frac{1}{.99}\]
\[\frac{15}{99}...\]

Answer 13

the formula for the sun of a geometric series is what you had already expressed is all

Answer 14

Nice job presenting the alternative method @amistre64.