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# STEP Prep Thread 2016 (Mark. II) watch

1. (Original post by drandy76)
Using the principles of strong induction (thank you zacin bas\$d God)
We can consider the initial case, where step has 3 letters.
1-S
2-T
3-E
4-P
.......... Oh my gawd, step 4 confirmed

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The 4th paper is called STEP S.

It's sat by applicants who want to skip Part IA so they can fasttrack the Tripos.

You need SSS in STEP I-III to qualify and the paper is taken in September.

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2. (Original post by jneill)
The 4th paper is called STEP S.

It's sat by applicants who want to skip Part IA so they can fasttrack the Tripos.

You need SSS in STEP I-III to qualify and the paper is taken in September.

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dont u mean...

STEPtember???
3. (Original post by drandy76)
Using the principles of strong induction (thank you zacin bas\$d God)
We can consider the initial case, where step has 3 letters.
1-S
2-T
3-E
4-P
.......... Oh my gawd, step 4 confirmed

Posted from TSR Mobile
Using proof by exhaustion:

Case 1: STEP is a subset of Stephen Siklos.
Case 2: STEP I, II, III are elements of the subset STEP.

Since STEP I, II, III are subsets of STEP, and due to the fact that STEP is a subset of Stephen Siklos, then there must be a distinct subset of elements of the set Stephen Siklos that are members of it. Therefore, using proof by exhaustion, the proposition that Stephen Siklos is hiding STEP IV is true. QED

My logic is irrefutable.

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4. (Original post by jneill)
The 4th paper is called STEP S.

It's sat by applicants who want to skip Part IA so they can fasttrack the Tripos.

You need SSS in STEP I-III to qualify and the paper is taken in September.

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I'm so excited to be the first half of the day before I get a follow back on my way home from work (spammed suggestive text and couldn't stop because sentence was interesting)

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5. (Original post by sweeneyrod)
dont u mean...

STEPtember???
That would be silly...

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6. (Original post by drandy76)
I'm so excited to be the first half of the day before I get a follow back on my way home from work (spammed suggestive text and couldn't stop because sentence was interesting)

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7. (Original post by Insight314)
Using proof by exhaustion:

Case 1: STEP is a subset of Stephen Siklos.
Case 2: STEP I, II, III are elements of the subset STEP.

Since STEP I, II, III are subsets of STEP, and due to the fact that STEP is a subset of Stephen Siklos, then there must be a distinct subset of elements of the set Stephen Siklos that are members of it. Therefore, using proof by exhaustion, the proposition that Stephen Siklos is hiding STEP IV is true. QED

My logic is irrefutable.

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Gonna take a page from my boy ramanujan and do a proof by intuition:
Consider how cool primes are(very)
Consider that they're are 3 step papers, a prime, it follows that Siklos would only work in prime number of step papers so there MUST be a FIFTH step paper too

Where's my field medal?

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8. (Original post by drandy76)
Gonna take a page from my boy ramanujan and do a proof by intuition:
Consider how cool primes are(very)
Consider that they're are 3 step papers, a prime, it follows that Siklos would only work in prime number of step papers so there MUST be a FIFTH step paper too

Where's my field medal?

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Using direct proof:

Axiom 1: Stephen Siklos is the set with subset of STEP.
Axiom 2: STEP is a subset with elements I, II, III; it's cardinality is a non-fixed positive integer, solely controlled by Stephen "the madman" Siklos.

By considering the above axioms, we can see that the cardinality of set Stephen Siklos is 13. Since 1 + 3 = 4, then through direct proof we have proven that the new cardinality of the set STEP, as outlined in Axiom 2, is one more than its cardinality over the real elements of the family. Therefore, there must be a STEP IV. QED

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9. (Original post by Insight314)
Using direct proof:

Axiom 1: Stephen Siklos is the set with subset of STEP.
Axiom 2: STEP is a subset with elements I, II, III; it's cardinality is a non-fixed positive integer, solely controlled by Stephen "the madman" Siklos.

By considering the above axioms, we can see that the cardinality of set Stephen Siklos is 13. Since 1 + 3 = 4, then through direct proof we have proven that the new cardinality of the set STEP, as outlined in Axiom 2, is one more than its cardinality over the real elements of the family. Therefore, there must be a STEP IV. QED

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One of these days I'm going to make a new TSR account called Stephen Siklos and post in this thread, to see what happens.
10. (Original post by Insight314)
Wait a minute.

1. STEP is made up of three letters.
2. STEP is also made up of three papers - I, II, III.

However, STEP is not made up of three letters, so by using proof by contradiction it has been proven that STEP is also not made up of three papers. STEP IV is being secretly hidden by Siklos confirmed. QED Illuminati controls STEP confirmed.

Try to refute my logic, you mortal peasants.

Pro-tip: You can't.

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Insight314's Lemma:

There always exists a bijection between the set of letters in the name of a set of exams and the set of the exams itself.
11. (Original post by Insight314)
Using direct proof:

Axiom 1: Stephen Siklos is the set with subset of STEP.
Axiom 2: STEP is a subset with elements I, II, III; it's cardinality is a non-fixed positive integer, solely controlled by Stephen "the madman" Siklos.

By considering the above axioms, we can see that the cardinality of set Stephen Siklos is 13. Since 1 + 3 = 4, then through direct proof we have proven that the new cardinality of the set STEP, as outlined in Axiom 2, is one more than its cardinality over the real elements of the family. Therefore, there must be a STEP IV. QED

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There are a bunch of STEP papers, possibly uncountably many,
Each step paper has a symmetrical domain about the point q=7, this each step paper can be formed by an odd and even function (but **** functions let's use numbers)
We know this is true for any step n where n is greater than or equal to one, by induction which I'm too lazy to type out

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12. I'm lost. What on earth is going on here.
13. What happened to vesniep

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14. (Original post by physicsmaths)
What happened to zacken

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dunno
15. (Original post by sweeneyrod)
STEP III 2015:
Spoiler:
Show
I thought this was a weird paper, due to the three questions on proof that weren't on any normal A-level stuff at all (not complaining, because they were quite nice).

2 - pretty straightforward
5 - also pretty straightforward, except I couldn't get the if part at the start of (ii)
6 - gave myself 2 marks for an attempt at half of (i)
7 - nice question
8 - would have been an easy full if I hadn't for some reason differentiated sin as -cos and vice versa. Instead it was an easy half.
12 - (i) fairly easy, although a bit dodgy that you didn't seem to have to prove that Rn's p.g.f. was always G(x). Didn't make much progress in (ii), and looking at the mark scheme it looked horrendous
13 - drew a little graph, gave up

Overall 73, a solidish 1. Not bad, although it was a bit odd I wouldn't mind a paper like it this year.
Should've attempted Q1, was lush (even though it took me ages lol). Well done anyway .
16. (Original post by drandy76)
My boy SS just dmed me, soz but you've been relocated to London met

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Hmm, not only do the first four letters of his name spell out STEP, his initials spell out the best STEP grades...
17. (Original post by IrrationalRoot)
Hmm, not only do the first four letters of his name spell out STEP, his initials spell out the best STEP grades...
Whoah, m8. What is SSS then?

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18. (Original post by Insight314)
Whoah, m8. What is SSS then?

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He's hinting that he couldn't care less about STEP I XD. (It's for all the other peasant universities lol.)
19. (Original post by Mathemagicien)
There is a STEP I?
Why did you get banned, naughty boy?

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20. (Original post by Mathemagicien)
I don't know
I am looking through your post history. I love the politically incorrect thread lol.

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