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# Will the real TeeEm please stand up! watch

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1. (Original post by Gilo98)
do 15 and 6 share a common power? on reflection that is perhaps a useless question....
clearly they do ...

you have to move away from whole numbers
2. (Original post by TeeEm)
clearly they do ...

you have to move away from whole numbers
Is it that obvious?

________________

15/2.5=6
square root of 6.25=2.5

??
3. (Original post by Gilo98)
Is it that obvious?
not at all...

it is a very hard AS question
(*****) in my scale.

Believe it or not there are some people here that find these questions too easy (because of their preparation for STEP/AEA/BMO) so I put the "extras" just for them.

Try it tomorrow when you are fresh
4. (Original post by TeeEm)
not at all...

it is a very hard AS question
(*****) in my scale.

Believe it or not there are some people here that find these questions too easy (because of their preparation for STEP/AEA/BMO) so I put the "extras" just for them.

Try it tomorrow when you are fresh
I will save it for saturday when I can sit down and do it properly.....will the solution be up by then? to think that by the end of the year I need to be finding this Q easy.....
5. (Original post by Gilo98)
I will save it for saturday when I can sit down and do it properly.....will the solution be up by then? to think that by the end of the year I need to be finding this Q easy.....
I put the solution up after 24 hours, so yes
6. (Original post by TeeEm)
not at all...

it is a very hard AS question
(*****) in my scale.

Believe it or not there are some people here that find these questions too easy (because of their preparation for STEP/AEA/BMO) so I put the "extras" just for them.

Try it tomorrow when you are fresh
Lol the extra is easier than the normal one
7. (Original post by Gome44)
Lol the extra is easier than the normal one
I wrote these questions in a hurry last night and the "normal" one is silly as it has a vital piece of information missing, though many of us might assume it.

here is the correct version.

Apologies
Attached Images
8. Q17, Thursday 16-07-15.pdf (12.9 KB, 52 views)
9. 27 days to A level results

Question for Friday 17/07/15
AS Maths Upwards
(a few messy surd questions coming up as I wrote 4 today)

and for those who want that little bit extra, a mathematical novelty
....
maybe not quite a question as it is very hard to see what to do, so maybe a bit of research to complete this proof
EDIT look for hint in post 790
Attached Images
10. Q18, Friday 17-07-15, solution.pdf (68.7 KB, 62 views)
11. Q18, Friday 17-07-15,.pdf (11.3 KB, 84 views)
12. Q18, Friday 17-07-15, EXTRA solution.pdf (149.7 KB, 63 views)
13. Q18, Friday 17-07-15, EXTRA.pdf (8.4 KB, 50 views)
14. (Original post by TeeEm)
27 days to A level results

Question for Friday 17/07/15
AS Maths Upwards
(a few messy surd questions coming up as I wrote 4 today)

and for those who want that little bit extra, a mathematical novelty
....
maybe not quite a question as it is very hard to see what to do, so maybe a bit of research to complete this proof
l acknowledge that when I awaken I shall firmly regret this decision. However, drunk me shall accept this next challenge. An inebriated mathematician is truly useful/incompetent.

morning edit: I don't recall typing this or seeing this question but as a mark of responsibility I shan't delete this post..for now.
Spoiler:
Show
For the extra, I have no concrete idea how to do this but...
For a non-terminating decimal expansion, which e of course has (I'm guessing this be assumed from the infinite series expression) can I say that all rational numbers with an infinite expression be expressed as the sum of one or more geometric series? i.e. 1/3 = 0.333333... so the sum of a geometric series with first term 0.3 and common ratio 1/10, 1/7 is 0.142857 recurring so we have the sums of 6 geometric series with common ratio (1/10)^6...
The expansion of e apparently doesn't fit these criteria as each term is multiplied by a new unique reciprocal (I mean, you first have 1 (well, 2 I guess, as you have 1 + 1 first), then 1/2, then 1/3, then 1/4 and so on due to the factorial stuff)

15. (Original post by 13 1 20 8 42)
l acknowledge that when I awaken I shall firmly regret this decision. However, drunk me shall accept this next challenge. An inebriated mathematician is truly useful/incompetent.

morning edit: I don't recall typing this or seeing this question but as a mark of responsibility I shan't delete this post..for now.
Spoiler:
Show
For the extra, I have no concrete idea how to do this but...
For a non-terminating decimal expansion, which e of course has (I'm guessing this be assumed from the infinite series expression) can I say that all rational numbers with an infinite expression be expressed as the sum of one or more geometric series? i.e. 1/3 = 0.333333... so the sum of a geometric series with first term 0.3 and common ratio 1/10, 1/7 is 0.142857 recurring so we have the sums of 6 geometric series with common ratio (1/10)^6...
The expansion of e apparently doesn't fit these criteria as each term is multiplied by a new unique reciprocal (I mean, you first have 1 (well, 2 I guess, as you have 1 + 1 first), then 1/2, then 1/3, then 1/4 and so on due to the factorial stuff)

This is the general structure

assume that e = p/q

p/q = 1 + 1 +1/2! + 1/3! + 4! + 1/q! + 1/(q+1)! +...

multiply by q!

move some terms on the left

LHS is a positive integer

so RHS is a positive integer

and somehow via inequalities and reasoning (usally found in mathematical analysis) the RHS is less than 1

so we found a positive integer less than 1

16. (Original post by TeeEm)
This is the general structure

assume that e = p/q

p/q = 1 + 1 +1/2! + 1/3! + 4! + 1/q! + 1/(q+1)! +...

multiply by q!

move some terms on the left

LHS is a positive integer

so RHS is a positive integer

and somehow via inequalities and reasoning (usally found in mathematical analysis) the RHS is less than 1

so we found a positive integer less than 1

There is a way to do it with floorbased functions.

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17. (Original post by physicsmaths)
there is a way to do it with floorbased functions.

posted from tsr mobile
there are several proofs

my question says "prove by contradiction"
18. I've noticed this thread has 40 pages. Sorry but what exactly is the purpose of this thread? I think we've established you're not Arsey
19. (Original post by kprime2)
I've noticed this thread has 40 pages. Sorry but what exactly is the purpose of this thread? I think we've established you're not Arsey
it is my office
20. (Original post by TeeEm)
it is my office
I see. So this is the place we come for your help?
21. (Original post by kprime2)
I see. So this is the place we come for your help?
No, but this is the thread on the background of my computer, when I am doing work in my actual office (except for another 2 weeks or so as I am on my holiday home in Greece).
22. (Original post by TeeEm)
there are several proofs

my question says "prove by contradiction"
This way is by contradiction. Im on holiday but ikl see if i can write it up later. It is beauftiful it was in a previous step question.

Posted from TSR Mobile
23. (Original post by physicsmaths)
This way is by contradiction. Im on holiday but ikl see if i can write it up later. It is beauftiful it was in a previous step question.

Posted from TSR Mobile
I will be truly interested to see it
24. (Original post by TeeEm)
No, but this is the thread on the background of my computer, when I am doing work in my actual office (except for another 2 weeks or so as I am on my holiday home in Greece).
Where in Greece out of curiosity?

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25. (Original post by aersh8)
Where in Greece out of curiosity?

Posted from TSR Mobile
It is in a nice strategic/geopolitical spot (best plot in Europe), alive, kicking and well.

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