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Need help

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Using infinite geometric series, show that 0.9 = 1

Mar 5, 2021

#1
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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
+159
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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