Express 2.909090909... as a Rational Number

Question

Put in the form $p/q$ with no common factors. My initial thought was to try to sum up the $.9090$ by noting $90/10^2$ repeats. However, I can't see what good that ultimately would do in finding the answer.

A general method for solving problems like this is found in the answers to http://math.stackexchange.com/q/295195/139123 — David K, Feb 22 '15 at 00:01

vanna · Accepted Answer · 2015-02-21T23:47:53.953

8

$$2.90909090... = 2+9\sum_{k=0}^\infty 10^{-(2k+1)} = 2+\frac{9}{10} \frac{1}{1-10^{-2}} = 2+\frac{90}{99}=\frac{32}{11}$$

edited Feb 21 '15 at 23:47

answered Feb 21 '15 at 22:32

vanna

1,514
8
21

The numerator and denominator are both divisible by $9$, so they can be reduced. – David K Feb 21 '15 at 23:38
True, i'll edit my answer thanks – vanna Feb 21 '15 at 23:47
That's good. In all fairness, though, $\frac{288}{99}$ is still a correct answer unless the problem says "a rational number in lowest terms." – David K Feb 21 '15 at 23:50
It does say that, just differently. – Jonathan Hebert Feb 22 '15 at 16:05

score 2 · Answer 2 · answered Feb 21 '15 at 22:30

2

$x=2.90909090...$ so $100x=290.90909...$, so $99x=288$. Hence $x=288/99$

answered Feb 21 '15 at 22:30

Robin Balean

678

score 2 · Answer 3 · answered Feb 21 '15 at 22:32

2

This is a neat trick that can also be used to show $0.99999... = 1$. Set $$x = 2.909090.... \quad \mbox{so that} \quad 100x = 290.909090...$$ Therefore, $$ 99x = 100x - x = 290.909090... - 2.909090... = 288, $$ so $2.909090... = x = 288/99$.

answered Feb 21 '15 at 22:32

Suugaku

2,469

score 1 · Answer 4 · answered Feb 21 '15 at 22:30

1

Let $x=2.909090\ldots$ and compute $99x = 100x-x$. This should be an integer (the fractional part is the same after multiplying by $100$, so $x$ is $\frac{\text{something}}{99}$.

answered Feb 21 '15 at 22:30

hmakholm left over Monica

286,031

score 1 · Answer 5 · answered Feb 21 '15 at 22:37

1

All repeating decimals with a zero in the one's place may be easily expressed as a fraction by putting the repeating sequence as the numerator, and a sequence of 9's in the denominator as long as the repeating sequence.

Thus, $0.90909090... = \frac{90}{99}$

So, we could quickly say that: $2.9090... = 2 + \frac{90}{99} = \frac{198}{99} + \frac{90}{99} = \frac{288}{99}$

answered Feb 21 '15 at 22:37

Jonathan Hebert

5,272

1

I always do it this way. – Karl Feb 22 '15 at 06:52

David K · Answer 6 · 2015-02-21T23:45:46.330

Let $x = 2.9090909\ldots.$ Then add $10x$ and $x$ as follows:

$$\begin{array}{rcr} 10x & = & 29.090909\ldots \\ +\ x & & +\ 2.909090\ldots \\ \hline 11x & = & 31.999999\ldots \end{array}$$

Therefore $11x = 32,$ since $0.999999\ldots = 1$ (exactly, not merely an approximation). Solve for $x$.

The answer above exploits the known fact that $\frac1{11} = 0.09090909\ldots.$ For a more general problem you can always get an answer by multiplying by $10^n - 1$ where $n$ is the number of repeating digits. In this case you would multiply by $99$, which happens to be $9\cdot11$, and you end up solving an equation like the one above except that both sides have been multiplied by $9$.

Express 2.909090909... as a Rational Number

6 Answers6