Will you please change the following repeating decimals to fractions:
1) 0.716 (1 is repeting) and 2) 4.23 ( 23 is repeating)
Thank you
In your first example I don't quite understand how the one can repeat if the 6 is after it, so I'll do the second one and hopefully you can use the method to figure out the first.
Let $\displaystyle x = 4.2323...$
Then $\displaystyle 100x = 423.2323...$
So, by subtraction, we have $\displaystyle 99x = 419$ (notice how the repeating cancels itself out when we subtract, this is very important)
So the fraction you want is $\displaystyle x = \frac{419}{99}$
It so happens that this is the same method used to prove that $\displaystyle 0.999... = 1$. Can you prove this as well?
I don't see why you have to bring out the geometric series to deal with something so trivial.
The algebraic 'proof' follows all the axioms, and so it is perfect in every respect; I don't see how you could dismiss it as a definition unless you are willing to dismiss the validity of algebra.
What axioms?
Ask thyself the question: "What does $\displaystyle .53535353....$ mean?".
You are assuming the following:
1)If you multiply the decimal by 10^k then you move the decimal k places.
2)If you add/subtract decimals you can do it componentwise (meaning first with first, second with second, ... )
To prove #1 and #2 you should use geometric series.
This is Mine 79th Post!!!
Thank you so very much for explaining this. I would think then the first problem should be 0.716 (716 repeating) and the fraction is 716/999
But how to solve it using the geometric series? because this is what we study now.
Thanks again.
Hello,
probably you have to do $\displaystyle 0.7161616161616... = \frac{358}{495}$
to #2 and the geometric serie:
$\displaystyle 4.23232323232323... = 4 + \frac{23}{100}+ \frac{23}{10000}+ \frac{23}{1000000}+ ...=$ $\displaystyle 4 + 23\left(\frac{1}{10^2}+ \frac{1}{10^4}+ \frac{1}{10^6}+ ...\right)$
I'm sure that you now easily see the geometric serie.