Which base of numerical system have $\frac 15 = 0.33333\ldots$?

Question

Which base of numerical system have $\frac{1}{5} = 0.33333\ldots$?

I need assistance in solving this one.

http://www.mathsisfun.com/binary-decimal-hexadecimal-converter.html is worth a look. It offers decimal, binary and hex conversions. It might help you see the relationships, ex, enter binary .010101010101 and see what this is in decimal. — JTP - Apologise to Monica, Jul 08 '14 at 18:28

score 28 · Answer 1 · answered Oct 29 '11 at 22:37

If we are working in base $b$ (we must have $b\gt3$), then $0.3333\ldots$ is $$0.3333\ldots = \frac{3}{b} + \frac{3}{b^2} + \frac{3}{b^3}+\cdots$$ Since $$\sum_{n=1}^{\infty}\frac{3}{b^n} = \frac{3}{b}\sum_{n=0}^{\infty}\frac{1}{b^n} = \frac{3}{b}\left(\frac{1}{1-\frac{1}{b}}\right) =\frac{3}{b-1},$$ then...

score 27 · Answer 2 · answered Oct 29 '11 at 22:36

27

Hint: if we multiply $0.33333\ldots$ by $5$ then we get $0.(15)(15)(15)(15)(15)\ldots$. Compare that to what happens when we multiply the same by $3$: $0.99999\ldots$, and its interpretation in decimal.

answered Oct 29 '11 at 22:36

Yuval Filmus

57,157

Gerry Myerson · Answer 3 · 2011-10-30T21:43:43.400

12

A reworking of Arturo's answer: let $x=0.333\dots$, let the base be $b$, then $$bx=3.333\dots$$ so $bx-x=(3.333\dots)-(0.333\dots)$ and you can take it from there.

edited Oct 30 '11 at 21:43

answered Oct 30 '11 at 02:22

Gerry Myerson

179,216

That should be $bx-x$, not $bx-b$. – jwodder Oct 30 '11 at 16:34
@jwodder correction made, thanks very much. – Gerry Myerson Oct 30 '11 at 21:44
So irrespective of the Base, the decimal shifts to the right when number is multiplied by the base? – MonK Jul 08 '14 at 19:42
@Sid, yes --- just think about what it means to write a number to a base. – Gerry Myerson Jul 09 '14 at 00:44

Which base of numerical system have $\frac 15 = 0.33333\ldots$?

3 Answers3