Mystery of Infinity-2

Number Theory Level 2

We all know the famous identity:

$0.9999\ldots{ }\ldots{ }\ldots = 1$ .

Here the numbers on the both sides are in base $10$ .

One special thing here about $9$ is: in this case, $9$ just one less than the base $10$ .

So, noticing this special property, one can make the conjecture that

The equality $0.ccc\ldots{ }\ldots{ }\ldots = 1$ is true in every base $b \geq 2$ , where $c$ is the largest digit in base $b$ , equivalently, $c=b-1$ .

Is this conjecture TRUE ?

Yes No

1 solution

Muhammad Rasel Parvej
Dec 2, 2016

In base $b\geq2$ , the left side can be written as

$c \times \frac{1}{b} + c \times \frac{1}{b^2} + c \times \frac{1}{b^3} + \ldots{ } \ldots{ } \ldots{ }$

= $(b-1) \times \frac{1}{b} + (b-1) \times \frac{1}{b^2} + (b-1) \times \frac{1}{b^3} + \ldots{ } \ldots{ } \ldots{ }$

= $(b-1) ( \frac{1}{b} + \frac{1}{b^2} + \frac{1}{b^3} + \ldots{ } \ldots{ } \ldots{ } )$

= $(b-1) \times \frac{\frac{1}{b}}{1-\frac{1}{b}}$ (as $b>1$ )

= $\frac{b-1}{b-1}$ , which is equal to $1$ for every $b\geq2$ .

Hence, Yes is the answer.

How did you pass from the infinite sum to (1/b)/(1-1/b)? Is it a known formula or is there any property there that made you get to that value?

Tomás Carvalho - 4 years, 6 months ago

Formula for sum of infinite geometric progression: $a+ar+ar^2+ \ldots{ }\ldots{ } \ldots{ }=\frac{a}{1-r}$ , with $|r|<1$ .

Muhammad Rasel Parvej - 4 years, 6 months ago

Mystery of Infinity-2

1 solution

0 pending reports