Simple question on Unending Decimals

[caudovittatus]

Hi I've decided to go back to the books and refresh my maths knowledge... I'm going over the introductory section of Engineering Mathematics by K. A. Stroud.

 

In the section on unending decimals he gives examples of how to convert an unending decimal into a fraction which goes like so (I haven't used his exact words below):

 

To convert 0.18181818 recurring into a fraction you multiply [math]0.\dot{1}\dot{8}[/math] by 100

 

So:

 

[math]

100 \times 0.\dot{1}\dot{8} = 18.\dot{1}\dot{8}

[/math]

 

Then if you subtract [math]0.\dot{1}\dot{8}[/math] from both sides you get:

 

[math]

100 \times 0.\dot{1}\dot{8} - 0.\dot{1}\dot{8} = 18.\dot{1}\dot{8} - 0.\dot{1}\dot{8} = 18

[/math]

 

That's fine so far, but he goes on (as printed in the Fifth edition):

 

"

That is:

 

[math]

99 \times 0.\dot{1}\dot{8} = 18.0

[/math]

 

This means that:

 

[math]

0.\dot{1}\dot{8} = \frac{18}{99} = \frac{2}{11}

[/math]

 

My problem is with [math]99 \times 0.\dot{1}\dot{8} = 18.0[/math] because I would have expected that to be [math]99 \times 0.\dot{1}\dot{8} \approx 18.0[/math] or am I just being dense? If you can explain why it is equal I'd greatly appreciate it.

 

Thanks in advance!

[XrenegadeZ]

Ignore my answer to your question if it doesnt meet your satisfaction. I am not a mathematician, just a curious reader on this board. I couldn't help but notice the two dots on the 0.18. I would say those dots would suggest number of units and as such, you are expected to give an approximation of the values you get, which is 18. so you are not wrong in my own point of view, but other mathematicians might give you a better response.

[DJBruce]

Let

 

[math]x=.\overline{18}[/math]

 

Now lets multiply each side by [math]100[/math]

so

 

[math]100x=18.\overline{18}[/math]

 

Now remember that:

[math]x=.\overline{18}[/math]

So lets subtract [math]x[/math] from each side:[math](100x-x)=(18.\overline{18}-.\overline{18})[/math]

So now we have:

[math]99x=18[/math]

This is where you got confused what you need to realize is that [math]x=.\overline{18}[/math] that means [math].1818181818[/math] forever so when I subtract [math]x[/math] from each side you are getting ride of all your numbers right of the decimal place.

so now that we have:[math]99x=18[/math]

We do a little simplifying and get

[math]x=\frac{18}{99}=\frac{2}{11}[/math]

If you want proof put [math]\frac{2}{11}[/math] into a scientific calculator and it will display [math].18181818[/math]. But the key is to remember that [math]x=.\overline{18}[/math] which gets rid of all the numbers on the right side of the decimal because [math].\overline{18}=18181818[/math] forever.

 

Also I believe the dots are suppose to be the bar which symbolizes that the decimal under the bar repeats infinity.

[XrenegadeZ]

I believe the poster wanted to find out why 99 x 0.18 was yielding 0.18 in the book when it should be yeilding 17.82. Was it an approximation or a bad calculation? Notice the approximate equality sign in his question... should it be equal to (=) or approximately (~) to?

Edited August 4, 2008 by XrenegadeZ

[Bignose]

 

My problem is with [math]99 \times 0.\dot{1}\dot{8} = 18.0[/math] because I would have expected that to be [math]99 \times 0.\dot{1}\dot{8} \approx 18.0[/math] or am I just being dense? If you can explain why it is equal I'd greatly appreciate it.

 

But 99x0.1818181... is equal to 18.0

 

You can see that from your own equations...

 

You agree that 2/11 is 0.1818181.... yes?

 

and 2/11 is equal to 18/99, right?

 

So 18/99 = 0.181818181818181818.....

 

Multiply both sides by 99 and you get

 

18 = 99*0.18181818181818181818........

 

You can do the limit, too.

 

99*(0.181818181...) = 99*(1/10 + 8/100 + 1/1000 + 8/10000 +...)

=99*(1/10) + 99*(8/100) + ....

 

Now let's look at the sum after each term

 

After 1 term, the sum is just 99(1/10) = 9.9

Add the next term, the 99*(8/100) and the sum becomes 17.82

after 3 terms: 17.919

4: 17.9982

5: 17.99919

6: 17.99998

 

I think you can see what the limit of these sums are. Each additional term brings the sum closer and closer to 18.

 

99*0.181818181818.... = 18.0

 

There is no approximation. It is an exact equation.

 

I believe the poster wanted to find out why 99 x 0.18 was yielding 0.18 in the book when it should be yeilding 17.82. Was it an approximation or a bad calculation? Notice the approximate equality sign in his question... should it be equal to (=) or approximately (~) to?

 

You forgot the infinitely repeating part

 

Sure 99*0.18 is 17.82

 

but 99*0.181818181818181818.... (where the ... denotes infinitely repeating) is equal to exactly 18.0

Edited August 4, 2008 by Bignose
multiple post merged

[ajb]

If you want proof put [math]\frac{2}{11}[/math] into a scientific calculator and it will display [math].18181818[/math].

 

Proof by calculator!

[XrenegadeZ]

Thanks Bignose for clarifying the significance of those two dots. I love these boards, dont know why i never signed up... ah well, at least i did.

[rafgas]

Hello all, im sorry for stealing this topic, but i ran into a problem on the exact same page, but my only problem is how come

 

So, same exact example. (i dont know how to do the points above the 18 to signify the repeating 18's. so ill just use the three dots to signfy repeating, even if its a little false).

 

100*0.18... = 18.18...

Subtracting .18... from both sides yields

 

100 * 0.18.. - 0.18... = 18.18... - .18...

that is

 

99 * .018... = 18.0

 

what? 0.18...-100 != 99 Where does the 99 come from? I see from the next few examples that it always seems to be 1000-0.0315... = 999. So i figure i could go ahead and just always go down one integer and be fine, but i want to understand it, if i dont understand, the chances of forgetting it are infinitely larger!

 

Anyway, thanks again, sorry for bad formatting of the numbers, and sorry for hijacking the thread.

/Rafgas

[The Observer]

100*0.18181818... - 1*0.18181818... = (100 - 1)*0.18181818... = 99*0.18181818...

[rafgas]

100*0.18181818... - 1*0.18181818... = (100 - 1)*0.18181818... = 99*0.18181818...

 

I dont understand, the 1*0.181818... isnt mentioned in the book. it goes straight from

 

100*0.18...=18.18...

 

"Subtracting 0.18... from both sides of the equation gives :

 

100*0.18... - 0.18... = 18.18...-0.18...

 

That is :

 

99 * 0.18... = 18"

 

I mean, the logic of your calculation seems correct, but why isnt it explained? it seems im missing something in either case.

So the step you added "1*0.18..." isnt in the book, but comes to the same conclusion. Not to mention, the book doesnt subtract 1 from each side, it subtracts 0.18...

 

0.18... - 100 is 99.818181... not 99

[imatfaal]

seems ok to me - longwindedly

[math] (100 * .\dot{1}81\dot{8}) = 18.\dot{1}81\dot{8} [/math]

 

less the repeated decimal from each side

 

[math] (100 * .\dot{1}81\dot{8}) - .\dot{1}81\dot{8} = 18.\dot{1}81\dot{8} - .\dot{1}81\dot{8}[/math]

 

is the same as

 

[math] (100 * .\dot{1}81\dot{8}) - (1 * .\dot{1}81\dot{8}) = 18.\dot{1}81\dot{8} - .\dot{1}81\dot{8}[/math]

 

play around with brackets

 

[math] (100 -1) * .\dot{1}81\dot{8} = 18[/math]

 

divide both sides by (100-1) or 99

 

[math] .\dot{1}81\dot{8} = 18/99 = 2/11 [/math]

[rafgas]

Thanks imatfaal.

 

I didnt realize that

 

means i can take one "unit" of .1818 away from the 100 units i started out with. so i felt very stupid for a while.

 

 

which is ofcourse what The observer tried to tell me, but i blocked on the "where did the 1*0.18 come from", >.<

Edited April 2, 2012 by rafgas

[imatfaal]

Thanks imatfaal.

 

I didnt realize that

 

means i can take one "unit" of .1818 away from the 100 units i started out with. so i felt very stupid for a while.

 

 

which is ofcourse what The observer tried to tell me, but i blocked on the "where did the 1*0.18 come from", >.<

 

Yep that's the step of importance - replace .1818... with x and it is easy, but with numbers our algebra goes awry and we want to do the sums too early