The Student Room Group

STEP Prep Thread 2016 (Mark. II)

Scroll to see replies

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


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.

Posted from TSR Mobile
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.

Posted from TSR Mobile


dont u mean...

STEPtember???
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.


Posted from TSR Mobile
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.

Posted from TSR Mobile


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)


Posted from TSR Mobile
Original post by sweeneyrod
dont u mean...

STEPtember???


That would be silly...

Posted from TSR Mobile
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)


Posted from TSR Mobile


That made so much sense!


Posted from TSR Mobile
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.


Posted from TSR Mobile


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?


Posted from TSR Mobile
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?


Posted from TSR Mobile


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





Posted from TSR Mobile
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


Posted from TSR Mobile


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.
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.


Posted from TSR Mobile


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.
(edited 7 years ago)
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





Posted from TSR Mobile


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


Posted from TSR Mobile
I'm lost. What on earth is going on here.
What happened to vesniep


Posted from TSR Mobile
Original post by physicsmaths
What happened to zacken


Posted from TSR Mobile


dunno
Original post by sweeneyrod
STEP III 2015:

Spoiler



Should've attempted Q1, was lush (even though it took me ages lol). Well done anyway :smile:.
Original post by drandy76
My boy SS just dmed me, soz but you've been relocated to London met



Posted from TSR Mobile


Hmm, not only do the first four letters of his name spell out STEP, his initials spell out the best STEP grades...
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?


Posted from TSR Mobile
Original post by Insight314
Whoah, m8. What is SSS then?


Posted from TSR Mobile


He's hinting that he couldn't care less about STEP I XD. (It's for all the other peasant universities lol.)
Original post by Mathemagicien
There is a STEP I? :biggrin:


Why did you get banned, naughty boy?


Posted from TSR Mobile
Original post by Mathemagicien
I don't know :frown:


I am looking through your post history. I love the politically incorrect thread lol.


Posted from TSR Mobile

Quick Reply

Latest