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Algebra Level 2

$\dfrac{0.999999\ldots}1 \, \boxed{\phantom0} \, 1$

What mathematical symbol should we fit inside the box above such that the equation/inequality holds true?

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3 solutions

Nihar Mahajan
Feb 7, 2016

If we have an infinite geometric progression with first term $=a$ and common ratio $=r$ then the sum of its infinite terms $=\dfrac{a}{1-r}$ . Hence we have:

$0.999\dots = \dfrac{9}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dots=\dfrac{\dfrac{9}{10}}{1-\dfrac{1}{10}} = \dfrac{\dfrac{9}{10}}{\dfrac{10-1}{10}}= \dfrac{\dfrac{9}{10}}{\dfrac{9}{10}} = \boxed{1}$

Only if r<1

A Former Brilliant Member - 1 year, 11 months ago

Sravanth C.
Feb 7, 2016

Let $0.999.... = x$ .

$10x = 9.999....\\x = 0.999...\\\therefore 9x = 9\implies \boxed{x=1}$

Hence $\dfrac{0.999...}1 = \dfrac 11 \boxed{=} 1$ .

We really have to set up a campaign for making newbies aware of the fact that $0.999...=1$ .

Swapnil Das - 5 years, 4 months ago

Yes, true!

Sravanth C. - 5 years, 4 months ago

There is more than what meets the eye when using such logic

T sidharth - 5 years, 3 months ago

Nice answer

Pablo Sanchez - 2 years ago

Surabhi Dixit
May 21, 2019

Let x = 0.999999........ (i)

Multiply eq. (i) by 10

10x = 9.999999..........(ii)

Subtract eq. (i) by (ii)

10x = 9.999999..........

-x = 0.999999..........

9x = 9

x = 9/9

x = 1

So, this justifies that 0.99999....... = 1

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