Question

$show \: that \: 0.3333... \infty \: can \: be \: expressed \: \\ in \: the \: form \: \frac{p}{q} \: \\ where \: p \: and \: q \: are \: \\ integers \: and \: q \: un = to \: 0.$
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Answer 1

QUESTION :-

★ Show that 0.3333.......∞ can be expressed in the form of p/q where p and q are integers q ≠ 0.

TO SHOW :-

• 0.3333.......∞ can be expressed in the form of p/q where p and q are integers q ≠ 0.

PROOF :-

$\\ : \implies \displaystyle \sf \: Let \: x = 0.33.... \infty = 0.\bar{3} \: \: \: \: \: \: \: \: \bigg \lgroup \: Equation \ 1 \bigg \rgroup$

$\dag \: \displaystyle \rm \:Multiply \: 10 \: on \: both \: the \: sides.$

$: \implies \displaystyle \sf \:10x = 3.333........ \infty$

$: \implies \displaystyle \sf \:10x = 3. \bar{3}\: \: \: \: \: \: \: \: \bigg \lgroup \: Equation \ 2\bigg \rgroup$

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$\\ \dashrightarrow \displaystyle \sf \bigg \lgroup \: Equation \ 2 \bigg \rgroup - \bigg \lgroup \: Equation \ 1 \bigg \rgroup$

$\dashrightarrow \displaystyle \sf10x = 3. \bar{3} \\ \displaystyle \sf \: ( - ) \: \: \: x = 0. \bar{3} \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \displaystyle \sf9x = 3.0$

$\dashrightarrow \displaystyle \sf9x = 3$

$\dashrightarrow \displaystyle \sf \: x = \dfrac{3}{9}$

$\dashrightarrow \underline{ \boxed{ \displaystyle \sf \: x = \dfrac{1}{3} }} \: \: \: \: \: \: \: \: \bigg\lgroup \displaystyle \sf\dfrac{p}{q} \: and \: q \neq0\bigg \rgroup$

$\therefore \underline{\displaystyle \sf \: 0.333..... \infty \: can \: be \: written \: as \: \dfrac{1}{3}}$

Answer 2

Given:

• 0.3333......$\infty$

To Prove:

• 0.3333......$\infty$ can be written in the form of p/q

Solution:

Here,

$\rm Let, x = 0.3333...........(i)$

$\underline{\boxed{ \color {green} \rm Now, multiplying \: eq(i) \: by \: 10}}$

$\rm \to10 \times x = 10 \times 0.3333.....$

$\rm \to10x= 3.3333........(ii)$

$\underline{\boxed{ \color {blue} \rm Now, subtract \: eq(i) \: by \: eq(ii)}}$

$\rm \longrightarrow 10x - x = 3.333.... - 0.3333$

$\rm \longrightarrow 9x= 3.333.... - 0.3333$

$\rm \longrightarrow 9x= 3 \cancel{.333....} - 0 \cancel{.3333.......}$

$\rm \longrightarrow 9x= 3$

$\rm \longrightarrow x= \frac{3}{9}$

$\rm \longrightarrow x= \frac{ \cancel3}{ \cancel9} = \frac{1}{3}$

$\rm \longrightarrow x= \frac{1}{3}$

Now, $\frac{1}{3}$ is a rational number in the form of $\rm\frac{p}{q}$ where, p and q are integers and q ≠ 0

$\underline{ \boxed{ \bold{Hence, Proved}}}$