Maths thread (1 Viewer)

drsabz101 · Jan 29, 2016

Q. How do I convert the decimal into a fraction :

0.2(re terminating)1(non-re-terminating)1 (re terminating)

InteGrand · Jan 29, 2016

drsabz101 said:
Q. How do I convert the decimal into a fraction :

0.2(re terminating)1(non-re-terminating)1 (re terminating)

$\noindent I don't understand what the number is. Is it: $0.21111...$ (i.e. $0.2\dot{1}$, also written as $0.2\bar{1}$)?$

drsabz101 · Jan 29, 2016

Okay its 0.2*11*

Asterix symbolizes the recurring dot. 1st number (2) reocurrs. 2nd number one (doesn't) the third (one) does

InteGrand · Jan 29, 2016

drsabz101 said:
Okay its 0.2*11*

Asterix symbolizes the recurring dot. 1st number (2) reocurrs. 2nd number one (doesn't) the third (one) does

$\noindent I'm still not sure what the number is. Maybe write it out a few times so we can tell the pattern. Is it $0.211211211211\ldots$?$

drsabz101 · Jan 29, 2016

i think it will turn out to be:

0.21121212121

drsabz101 · Jan 29, 2016

I would take a photo. But it's my study period and I can't use my phone

BLIT2014 · Jan 29, 2016

drsabz101 said:
I would take a photo. But it's my study period and I can't use my phone

http://www.basic-mathematics.com/converting-repeating-decimals-to-fractions.html

InteGrand · Jan 29, 2016

drsabz101 said:
i think it will turn out to be:

0.21121212121

$\noindent Let $\chi=0.211212121\ldots$. Then $\chi=0.211+0.000212121\ldots=0.211+x$, where $x=0.000212121\ldots$.$

$\noindent Note that $x=10^{-3}\cdot y=\frac{1}{1000}\cdot y$, where $y=0.212121\ldots$. Now, $100y=21.21212\ldots$. So $100y-y =\left(21.21212\ldots\right) - \left(0.212121\ldots\right) = 21\Rightarrow 99y=21\Rightarrow y = \frac{21}{99}=\frac{7}{33}$. So $x=\frac{1}{1000}\times \frac{7}{33}=\frac{7}{33000}$. Now, since $0.211 = \frac{211}{1000}$, we have$

$\begin{align*}\chi&=\frac{211}{1000} + x \\ &= \frac{211}{1000} + \frac{7}{33000} \\ &= \frac{211\times 33 + 7}{33000} \quad (\text{taking a common denominator}) \\ &= \frac{6970}{33000} \\ &= \frac{697}{3300}.\end{align*}$

drsabz101 · Jan 29, 2016

I know the method. But this time the first repeats and the last number repeats, not the middle which is confusing

InteGrand · Jan 29, 2016

drsabz101 said:
I know the method. But this time the first repeats and the last number repeats, not the middle which is confusing

For these types of Q's, you can do it as I did above (separate out the decimal so it is a finite pattern + simple recurring pattern).

An alternate way of converting a simple recurring pattern decimal to a fraction is by using geometric series.

drsabz101 · Jan 29, 2016

Hey Integrand. thanks for the reply, I am not getting what happened tin the first 2 lines, where you split a then substitued the value of x. How did you get the value of x to be 0.000212121..

InteGrand · Jan 29, 2016

drsabz101 said:
Hey Integrand. thanks for the reply, I am not getting what happened tin the first 2 lines, where you split a then substitued the value of x. How did you get the value of x to be 0.000212121..

$\noindent This is because$

$\begin{align*}0.21121212121\ldots &= 0.211 + 0.000212121\ldots \quad (\text{separate out the first three decimal places. It's like how }0.56414 = 0.564 + 0.00014).\end{align*}$

drsabz101 · Jan 29, 2016

Ohhh okay I get it. Thankyou!!

drsabz101 · Jan 29, 2016

Another q. for the section 2 - where did y come in?

InteGrand · Jan 29, 2016

drsabz101 said:
Another q. for the section 2 - where did y come in?

$\noindent We had a number like $x=0.000212121\ldots$, so since it's nicer to work with the decimal with the first non-zero decimal place started right after the decimal point, we rewrote that as $x=10^{-3}\cdot \underbrace{0.212121\ldots}_{\text{call this } y}$ (i.e. shift the decimal place in $x$ three spaces to the right and compensate by dividing by $10^3$, which is equivalent to multiplying it by $10^{-3}$).$

Drsoccerball · Jan 29, 2016

InteGrand said:
$\noindent Let $\chi=0.211212121\ldots$. Then $\chi=0.211+0.000212121\ldots=0.211+x$, where $x=0.000212121\ldots$.$

$\noindent Note that $x=10^{-3}\cdot y=\frac{1}{1000}\cdot y$, where $y=0.212121\ldots$. Now, $100y=21.21212\ldots$. So $100y-y =\left(21.21212\ldots\right) - \left(0.212121\ldots\right) = 21\Rightarrow 99y=21\Rightarrow y = \frac{21}{99}=\frac{7}{33}$. So $x=\frac{1}{1000}\times \frac{7}{33}=\frac{7}{33000}$. Now, since $0.211 = \frac{211}{1000}$, we have$

$\begin{align*}\chi&=\frac{211}{1000} + x \\ &= \frac{211}{1000} + \frac{7}{33000} \\ &= \frac{211\times 33 + 7}{33000} \quad (\text{taking a common denominator}) \\ &= \frac{6970}{33000} \\ &= \frac{697}{3300}.\end{align*}$

lol integrand man what are you doing you made it look so hard ahhaha

$x= 0.211212121$

$100x=21.12121212..$

$10000x=2112.12121212$

$(10000-100)x=2112-21$

$x=\frac{697}{3300}$

Green Yoda · Jan 29, 2016

lel we finished basic arithmetic and algebra in 2 days (1.5 actually cuz we spent half a lesson getting textbooks)

InteGrand · Jan 29, 2016

Drsoccerball said:
lol integrand man what are you doing you made it look so hard ahhaha

$x= 0.211212121$

$100x=21.12121212..$

$10000x=2112.12121212$

$(10000-100)x=2112-21$

$x=\frac{697}{3301}$

(Btw you did a typo, it should be 3300 in the denominator.) I did do it correctly, but yeah, your method is faster.

si2136 · Jan 29, 2016

Finally. I can practice some maths now.

drsabz101 · Jan 29, 2016

why couldn't you before?

lol don't worry I get what you mean

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