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Using infinite geometric series, show that 0.9 = 1

 Mar 5, 2021
 #1
avatar+128 
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0.9=9/10.

I think can be like this:

\(0.999...=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...\)

\(0.999...=​​\frac{9}{10}+(\frac{9}{10})(\frac{1}{10})+(\frac{9}{10})(\frac{1}{10})^2+...\)

Using infinite geometric series, with a=\(\frac{9}{10}\) and r=\(\frac{1}{10}\):

\(0.999...=\frac{\frac{9}{10}}{(1-\frac{1}{10})}=\frac{9}{10}\times\frac{10}{9}=1\)

Hope this help.

 Mar 5, 2021
 #2
avatar+159 
0

Nice Answer James!
But, I would like to think of it a different way.

0.9 cannot be equal to 1 only \(0.\overline{9}\) can. If you are meaning\(0.\overline{9}\) then, I can show you how to do it.

1/9 = \(0.\overline{1}\)

2/9 = \(0.\overline{2}\)

3/9 = 1/3 = \(0.\overline{3}\)

And so on this means that when we come to 9/9 = \(0.\overline{9}\)=1

🐺🐺🐺

vockeyvockvock  Mar 5, 2021

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