0.34545... decima to fraction

Question

this is my assignment and i dont know how to answer.

Answer 1

Look for the part that repeats, the recurring decimal. We need to separate the fixed part from the recurring part. In this case, multiply the decimal by 10 to give 3.4545... The recurring part is 45. Divide this 2-digit number by the number with as many 9's: 45/99=5/11. When we separated the decimal we had 3.45... We can now replace the recurring decimal with 5/11. Make it into an improper fraction by multiplying the integer 3 by the denominator 11 and adding the numerator 5. We divide the result by the denominator 11 giving us 38/11. But we have to compensate for multiplying by 10 to separate the two parts of the decimal, so now we divide this fraction by 10: 38/110, which cancels down to 19/55. Check this out on a calculator and you'll see it's 0.34545...

Answer 2

Given: 0.3454545...   Assuming the number is a recurring decimal that repeats "45" forever, we try to find ea equivalent fraction.

Let A=0.3, and B=0.0454545..., so 0.3454545=A+B.

Change both decimals,A and B, into fraction forms separately.  

A=0.3=3/10, and B can be changed in ways shown below:

Sol.1: B=0.0454545...=0.045+0.00045+0.0000045+...

=45(0.001+0.00001+0.0000001+...)=45x0.0010101...  

While, 1/99=0.010101..., and 1/990=0.00101010...,

so B=45x1/990=5/110.   Thus A+B=3/10 + 5/110=38/110=19/55

The answer is: 0.3454545...=19/15

 

Sol.2: 1000B=45.4545..., and 10B=0.4545...,

so 1000B-10B=45, that is 990B=45.  

We have: B=45/990=5/110   Thus, A+B=3/10 + 5/110=19/55

The answer is: 0.34545...=19/55

 

Sol.3: B=0.045+0.00045+0.0000045=45/10^3(1 + 1/10^2 + 1/10^4 + 1/10^6...)

Use the formula of sum,S, for infinite geometric series, where the first term is a and the common ratio is r (lrl<1): S=a/(1-r)   Here, a=1 and r=1/10^2

So, B=45/10^3 x 1/(1-1/10^2)=45/10^3 x 10^2/(10^2-1)=45/990=5/110  

Thus, A+B=3/10+5/110=19/55

The answer is: 0.34545...=19/55