Homework question

[xud]

My maths is not what it used to be. Need some assistance.

Q: Prove that the recurring decimal 0.25 = 25/99


Thanks

Mike

[Nibble]

25 divided by 99 = 0.252525~~~~

[bamberger]

Nibble -I don't see how that is a proof.

0.252525.... = 0.25 +.25x.01+ ..25 x .0001 etc

So we have a geometric series and the sum is .25 x 1/1-.01=.25/.99=25/99

[Hymie]

I like Nibble’s answer.

Without too much difficulty (based on 25/100 = 0.25), you should be able to show that 25/9900 = 0.002525252525….

But even that does not ‘prove’ 25/99 = 0.2525252525….., such ‘proof’ would probably win you a Nobel prize.

[Hymie]

Type Your Answer Here...To claim my Nobel prize (and get you full marks for your homework).

25/100 – 25/99= 0.0025252525…….

Therefore (2500 – 2475)/9900 = 0.0025252525…..

25/9900 = 0.0025252525….

or even 5/1985 = 0.0025252525…

or even 1/396 = 0.0025252525…

Time to get to the pub.

[Hymie]

I should add that Nibble's answer is just as valid as mine (if not more so).

I'm getting my coat now.

[bamberger]

Hymie Where do you get 25/100 – 25/99= 0.0025252525……. from?

[Jonathan-Joe]

In Nibble's answer, it is only by inference that we see that the division 25/99 poduces an infinitely recurring decimal and is therefore not a proof. Bamberger's proof seems incomplete and Hymie's, if correct, seems long winded.

The following is the classic way of proving this:

Let X = .2525252525... (1)
then 100X = 25.2525252525... (2)
Subract equation (2) from equation (1)
99X = 25, so X = 99/25

This works because we can see that for any position after the decimal point, the digit in each expression (1) and (2) is equal, so subtraction of one from the other leaves zero after the decimal point.

This proof can be used to find a fraction equal to any recurring decimal, using higher powers of 10 to match length of the repeating decimal. Where we used 100 above for .25 recurring, use 1000 for .258, 10000 for .2587 etc

Prove .25872587 recurring = 2587/99999
Let X = .25872587... (1)
then 10000X = 2587.25872587... (2)
Subtract (1) from (2)
9999X = 2587, so X = 2587/9999

[Jonathan-Joe]

I tried to put spaces in my equations, but AB removed them. If you write the equations with dcimal points aligned you will immediately see how this works.

[factor-fiction]

I wish I'd seen this thread earlier.
Jonathon-Joe's approach is the best way to prove these- if you are doing higher GCSE Maths that's the way you're expected to do it

[bamberger]

I don't see why my proof is in any way incomplete. Kindly explain .

[factor-fiction]

bamberger's proof is fine also, but is more appropriate for first year A level and assumes a knowledge about the sum of a geometric series.

[bamberger]

Good grief times change -we did geometric progressions in the third year of senior school!

[factor-fiction]

Hi bamberger- I bet, like others of our age, you did integration and volumes of revolution at 'O' level too but students don't touch calculus until A level now. However we didn't do any probability at O level, I'm sure, whereas it's taught much earlier now.

[Jonathan-Joe]

Yes, it's the jump to "... the sum is ... ". You know how to sum a geometric progression but you can't assume the OPer does - for the level of this question, your method needs a line or two showing the steps to build the expression "25 x 1/1-.01".

[bamberger]

I'm also intrigued as to why and when pupils became students . Yes we differentiated and integrated (except 1/x) -probability and stats weren't on the syllabus .

[factor-fiction]

Many schools now call themselves Colleges or Academies, and the word student is generally preferred to pupil.
Teaching is now called Teaching and Learning
At some school the heads are called the Lead Learner!
Departments are now called schools/colleges or even 'learning villages' i noticed in one advert

[heathfield]

And if you look online for past papers, despite all the denial of 'dumbing down', you'll find the same questions which were in the 1952 Eleven Plus exams are these days included in GCSE papers.

[factor-fiction]

hi heathfield- I think you are referring to the current Foundation level papers. These give grades of U, G, F, E, D or C. They are equivalent to the old CSE (for which a grade 1 was considered to be worth an 'O' level grade c/6. These papers do start with easier questons, some of which will indeed be similar to old 11 plus Maths questions
To find the equivalent of 'O' levels you need to look at Higher tier GCSE Maths papers. The bulk of what you and I covered at O level will still be on there, although a few newer topics are now on there (e.g. frequency densities, box plots) in place of a few things such as calculus. These papers bear little resemblance to old 11 plus questions

[vinginge]

Let the original recurring decimal be n.
Multiply the recurring decimal by 100. It becomes 25.2525....
Subtract the orginal recurring decimal...................0.2525....
And the difference is 25 = 100n - n = 99n
Thus, n = 25/99
Hope I'm not too late and apologies if anyone else has already provided this solution.

[factor-fiction]

Yes, good solution vinginge.
Jonathan-Joe beat you to it by a few days but yours is correct whereas J-J's contained a typo (where he said X = 99/25, he meant X=25/99)