Question

show that 2.999=2.9 can be expressed in the form p/q where p and q are integers of q is not equal to 0
​

Answer 1

Let,

$\displaystyle\longrightarrow\sf{x=2.999...\quad\quad\dots (1)}$

Multiplying by 10,

$\displaystyle\longrightarrow\sf{10x=10\times2.999...}$

$\displaystyle\longrightarrow\sf{10x=29.999...\quad\quad\dots (2)}$

Subtracting (1) from (2),

$\displaystyle\longrightarrow\sf{10x-x=29.999...-2.999...}$

$\displaystyle\longrightarrow\sf{9x=27}$

$\displaystyle\longrightarrow\sf{x=\dfrac {27}{9}}$

$\displaystyle\longrightarrow\sf{x=3\quad\quad\dots (3)}$

From (1) and (3),

$\displaystyle\longrightarrow\sf{2.999...=3}$

3 is a rational number and can be written in the form $\displaystyle\sf {\dfrac {p}{q}}$ for some integers $\displaystyle\sf {p}$ and $\displaystyle\sf {q}$ and $\displaystyle\sf {q\neq0.}$ So can be $\displaystyle\sf {2.999...}$

Hence Proved!

Answer 2

Answer:

• Let us assume x as 2.999...

• Here we see the digits are non-terminating recurring that's the repeating decimal expansions.

Now, we multiply x by 10, to get:

= 10x = 2.999...

= 10x = 10 * 2.999...

= 10x = 29.999...

= 10 (x - x) = (29.999... - 2.999...)

Now, x is equal to 27.

Expressing in the form of p/q where q ≠ 0:

= 27/9

= 3

.•. 3 = 2.999

Therefore, 2/9 is the required form of p/q and 3 and q ≠ 0.