Proof:
lim(m --> ∞) sum(n = 1)^m (9)/(10^n) = 1
0.9999... = 1
Thus x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.
Printable View
Proof:
lim(m --> ∞) sum(n = 1)^m (9)/(10^n) = 1
0.9999... = 1
Thus x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.
It's one of the first things you learn when dealing with calculus. I find it rather logical.
I mean, let's divide 1 by 3.
1/3 = 0.333...
Multiply it by 3 and you should be back to 1:
3 * 1/3 = 3 * 0.333... = 0.999... = 1
---
Oh yeah, this is SB. My post needs some senselessness or it'll be kicked out:
Whenever green fairies blink, the rush of adrenaline makes half-full cups of tea fly aimfully to the right ear of the person sitting by the door.
We've been through this.
[Edit] All of what I said was false. Joke's on me.
Invader, just go ask any maths teacher. 0.999~ equals one. Proof is on the first post. You forget that the nines go on endlessly. The limit of a number *is* the number.
Today, you buy a cheese. Tomorrow, you will buy half a cheese. After tomorrow, you will buy half of half of a cheese, and so on. If you continue to do this endlessly, how much cheese will you have at the end? Answer is 2 whole cheeses. Not nearly 2 whole cheeses, but 2 whole cheeses exactly.
---
0.333 is not equal to one third, but the repeating decimal 0.3333333~ does equal 1/3.
fixed. All that stuff that comes after is unnecessary. the fact is that every real number is defined to be an infinite sum as above. So the left hand side is just the definition of .99999...... If you don't want to parse that, it's just .9 + .09 + .009 + .0009 + .......Quote:
Originally Posted by hellohihello
The value of this infinite sum is defined to be the limit of its sequence of partial sums. You get the i'th member of this sequence by adding the first i terms of the infinite sum. so it is {.9, .99, .999, .9999, ..... } the limit of this sequence is 1 so by definition, the value of the infinite sum is one and again, by definition, the value of .99999...... is 1. Opposing arguments RIP.
@invader
I think you meant hyberbolas and that can't be represented by an elementary function in the way that I think you are thinking, it can be represented as a vector valued function or as the set of zeros of a polynomial. If we adjoin the "line at infinity" and pass to projective coordinates, then it does have a value at infinity, namely 0, and is really just an elipse that intersects the line at infinity in two places. The fact that it doesn't have a value at infinity is really just a defect of the non-projective coordinates. A parabala, in that case also has a value at infinity and is just an ellipse that intersects the line at infinity twice at the same place in the same way that (x - 0)^2 = x^2 intersects the x axis twice at 0.
But it is exactly why 0.999~ does not equal 1.Quote:
Originally Posted by PhilosopherStoned
It is used for all intents and purposes to equal 1.
But by its very definition it "infinitely approaches 1, never reaching it".
So I don't get why this whole debate ever comes to be.
0.999~ is just that. And 1 is just that.
They are not equal, but the former is rounded to the latter.
Because during application the difference is negligible.
However, when you say "something is equal to something else", that isn't about application.
It's about static conditions.
So when saying 0.999~ is equal to 1, you're saying two different values (no matter how small the difference is) are the same.
And that's just obviously bogus.
In other words, you don't even have to know any math or anything about parabolas or any such to realise that if two different values are defined to be different, they are different. Doesn't matter how they are applied in calculations and all. Conceptually (not mathematically), they are different.
.Quote:
Originally Posted by Black_Eagle
two different contexts.Quote:
Originally Posted by Merlock
You're confusing the definition of an infinite sum with an infinite sequence.Quote:
Originally Posted by Merlock
because people that know nothing about math like to feel like they have any input to offer.Quote:
Originally Posted by Merlock
precisely. They are both equal to 1.Quote:
Originally Posted by Merlock
See above. Is there something you don't understand? I would be more than willing to help you.Quote:
Originally Posted by Merlock
from a formal perspective, all mathematics is about static conditions..Quote:
Originally Posted by Merlock
see the proof in my previous post.Quote:
Originally Posted by Merlock
reread my post. .9999...... is defined to be the limit of the sum which is 1. You don't need to know anything about math but it helps if you want to understand stuff. This is math, not literature. There is one truth value that can be assigned to any statement.Quote:
Originally Posted by Merlock
Hope that clears stuff up.
It doesn't clear stuff up, since you yourself agreed with my stance.Quote:
Originally Posted by PhilosopherStoned
You agree that it's a fact that there is no value for it at infinity.Quote:
Originally Posted by PhilosopherStoned
No value. Not 1.
And if it isn't 1, it isn't equal to 1.
Doesn't matter how it's applied. Conceptually, it never reaches 1.
Quote me agreeing with your stance. Do you mean about it being static? you're right. That has no bearing on the question at hand.Quote:
Originally Posted by Merlock
I was talking about an algebraic curve which is a function. It was an aside for invader. It's a cool fact. This is totally different.Quote:
Originally Posted by Merlock
Here's another way to think about it from a topological perspective. The real numbers form a continuum. This means that given two real numbers, a and b, I can always find a third real number c, such that a < c < b. I can always stick a number in between them. All you have to do to prove that .9999...... is not equal to 1 is find some number that you can stick between the two.
here's yet another way to think about it. What's 1 - .99999999.........?
essentially what you are doing is saying that 4 + 1 =/= 3 + 2 because I am writing them differently. 1 = 1.000000....... = 1 + .0 + .00 + .000 + ....
is an infinite sum as well. It just has trivial terms after the first but is really just another way of writing .99999..... = 0 + .9 + .99 + .999 + .....
This is what I meant when I said I don't understand why this whole debate comes up.Quote:
Originally Posted by PhilosopherStoned
What you said in the above quote has no relevance whatsoever.
There is simply no need to prove or disprove that 0.999~ = 1.
By definition, 0.999~ infinitely reaches towards 1 but never gets there.
And calculations like those limits up there or formulating things like a < c < b...
None of it matters.
If something is defined specifically to never be equal to something else...it isn't.
In other words, what I mean to say is that I don't see any reason for any of the above proof to even exist.
And as such, I shall withdraw from this topic so as to avoid taking part in trying to prove or disprove something, heh. Just wanted to get my position out there, as it's always misinterpreted as me trying to disprove something. When in fact, I fully realise that 0.999~ and 1 are two different values by definition, and I simply wonder where anyone got the idea to start a debate about their equality.
It's like someone seeing 5 apples lying near a pile of 476 metal bolts, then suddenly getting up and stating, "I'm going to prove that 5 = 476!" :P
To all of you who can't understand how .999~ = 1, keep in mind you can't argue math in words.
So, provide a formal proof showing .999~ =/= 1 or GTFO.
Edit: Also, I find it hilarious that this was re-posted in Senseless Banter.
Ahrgh!
These threads are like an infinitely recurring nightmare.
NO.Quote:
The .9999~ is not equivalent to the value that comes after the decimal in 9.9999~.
NO.Quote:
But it is exactly why 0.999~ does not equal 1.
It is used for all intents and purposes to equal 1.
But by its very definition it "infinitely approaches 1, never reaching it".
Numbers don't APPROACH anything, they are FIXED values. Note that 0.999~ represents an INFINITE string of 9s, not ROUGHLY 0.999.
The fact is that it has been proved in the first post and you can't argue, as has been said above, with a mathematical proof. Especially not with words and analogies.
Lord, you sound extremely intelligent. I'm in awe at the largeness of your words.Quote:
it's widely accepted in mathematics that the function that represents a parabola can approach but never be equal to it's asymptote at any point on the graph, even though it can close the 'space' to infinitesimally small distances.
Except parabolas don't even have any asymptotes.
Xei speaks truth.
http://i202.photobucket.com/albums/a...sShitAgain.jpg
We had two threads going for a while where we discussed this. My conclusion was that the two figures are equal and that it has been proven but that it has never been fully explained how the 9's go on infinitely yet comprise a value of 1, even though calculus says they do and there are terms for the phenomena involved. I know the 9's are not moving, but they do "go" on forever. There is no end to infinity. The situation seems paradoxical.
To prove it in a different way, you can actually use the formula for calculating the sum of an infinite geometric progression. The initial term would be 0.9 and the ratio would be 0.1:
S = (0.9 + 0.09 + 0.009 + ....)
S = a1 / (1 - q)
S = 0.9 / (1 - 0.1)
S = 0.9 / 0.9
S = 1
You deny Blizzard? They will ban you from the internet.
It really comes down to math vs. logic. Math tells us that it does equal 1, but logic tells us just the oposite.Quote:
Originally Posted by Universal Mind
I think I've figured out how to use what Kromoh posted to argue that .999~ is not equal to 1. I'm sure my argument will be immediately proven wrong by someone far better at math than me, but I think I'll post it anyway, just for the hell of it. I mean, this is SB after all.
It is generally accepted that 1/3=.333~. So, if 1/3x3=1, then .333~ should also equal one. But it comes out to .999~. This would seem to say that .999~=1. However, consider this: no matter how many 3's you add on to that .333~, it is never exactly equal to 1/3. Therefore 1/3≠.333~, meaning that 1/3x3≠.333~x3, and .999≠1.
I think that the key to grasping it intuitively is to consider the quantity 1 - .999~
If we can show that 1 - .999~ = 0 then this is equivelent to saying that 1 = .999~
Lets go ahead and say that d = 1 - .999~ just so we don't have to write it out every time.
There is a theorem that says that if x is a number and x is not negative and x is less than any positive real number, then x = 0. Does that make sense?
It should be obvious that d is not negative because for 1 - b to be negative, b has to be bigger than 1. But .999~ is at the very least not bigger than 1 so d is not negative.
next, we need to prove that d is smaller than any positive real number. We can then apply the above theorem to conclude that d = 0.
We will do this by approximating d. Let b(1) = 1 - .9 = .1, b(2) = 1 - .99 = .01 and so forth.
For example, b(14) = 1 - .99999999999999 = .00000000000001. The bs are all exact numbers and each one is closer to d than the last one so they are good approximations.
Also, each b is larger than d. So given a positive real number, simply choose a number j so that b(j) has more zeros after the decimal place than the number in question. It is then smaller then the number in question but still larger than d. For example, given the number .00005, just notice that it is non-zero in the fifth decimal place and take b(5) = .00001 < .00005. This shows that d is smaller than that number. Because all we assumed was that the number is positive, we can do this for any positive number, however small. We can then apply the theorem to conclude that d = 0.
This means that d = 1 - .999~ = 0 so 1 = .999~
The word you're looking for is not logic, but intuition. Mathematics is the very essence of logic.Quote:
Originally Posted by Supernova
Hmm, to use your thought stream:Quote:
I think I've figured out how to use what Kromoh posted to argue that .999~ is not equal to 1. I'm sure my argument will be immediately proven wrong by someone far better at math than me, but I think I'll post it anyway, just for the hell of it. I mean, this is SB after all.
It is generally accepted that 1/3=.333~. So, if 1/3x3=1, then .333~ should also equal one. But it comes out to .999~. This would seem to say that .999~=1. However, consider this: no matter how many 3's you add on to that .333~, it is never exactly equal to 1/3. Therefore 1/3≠.333~, meaning that 1/3x3≠.333~x3, and .999≠1.
assuming 1/3 = .333~
3 * 1/3 = 3 * 0.333~
1 = 0.999~
The matter is that it's not an arbitrary amount of 3's you put there. Meaning, you don't just sit down in a table and start writing 3's until you get tired. 0.333~ is nothing more than the decimal representation of the rational number 1/3. The amount of 3's contained in 0.333~ is infinite. That is the reason why 0.333~ = 1/3.
0.333~ is not roughly 0.333, it is the decimal representation of the number 1/3, a so-called repetitive decimal.
The paradox of the cheese illustrates this well. It's in a previous post of mine somewhere.
Sorry for the double post but Supernova posted this when I was working on my last one.
It comes down to math vs. naive intuition. naive is not offensive in this context, it just means what you would think if you weren't a mathematician. Most mathematicians use "naive set theory" for instance, because it's good enough for their needs.Quote:
Originally Posted by Supernova
Great. It's how one learns how to construct a proof ;)Quote:
Originally Posted by Supernova
I bolded and blued the problem area. The question of if .333 = 1/3 is really the same issue as if .999~ = 1 or 5.999~ = 6 for that matter. The fundamental question is if one number can have multiple decimal representations. It might help to remember the definition of a rational number (a fraction) as a number with infinitely repeating decimals. .333~ is the infinitely repeating decimal expansion of the rational number 1/3, .999~ is the decimal expansion of 1/1 = 1 and so on.Quote:
Originally Posted by Supernova
Well I guess you've only proven that I don't really get this kind of advanced math, something I could have told you myself :P
(yea, maybe logic isn't the best word for what I was trying to describe)
I demand moderator powers so that I may lock this abomination of a thread.
Well then don't read it. You might as well lock every thread about athiesm vs. god. at least this one has a definitive, absolute answer with proof and everything. quit your bellyaching.Quote:
Originally Posted by Black_Eagle
Why is this in SB anyway? This thread deserves to die in a boring place like Science & Maths, not in a lively place such as the SB.
Yesh that's a nice way, or you can even go back to the basics of proving the GR formula:Quote:
To prove it in a different way, you can actually use the formula for calculating the sum of an infinite geometric progression. The initial term would be 0.9 and the ratio would be 0.1:
S = (0.9 + 0.09 + 0.009 + ....)
S = a1 / (1 - q)
S = 0.9 / (1 - 0.1)
S = 0.9 / 0.9
S = 1
S = 0.9 + 0.09 + 0.009 + ...
0.1S = 0.09 + 0.009 + 0.0009 + ...
S - 0.1S = (1 - 0.1)S = 0.9S = 0.9
=> S = 1
As Kromoh said, you're talking about intuition, and the problem is that intuition is subjective, unlike mathematics. As a matter of fact I find it totally intuitive that 0.999... is equal to 1, as the number is infinitely close to 1 and hence equal; I do not understand anybody who finds the opposite intuitive.Quote:
It really comes down to math vs. logic. Math tells us that it does equal 1, but logic tells us just the oposite.
That's why we have mathematical proof anyway. Sometimes results are surprising.
I wasn't expecting serious answers in SB -_-
0.999~ = 0.9 actually. :V
I may be guilty on that one +.+Quote:
Originally Posted by hellohihello
You can't comb the hair of a spherical dog without introducing a part
When I saw the post I though "well, that's all wrong. It can't be"
Then I actually read some of the posts, and it makes perfect sense. Why do people continue to deny it? >.< :cheeky:
The complex numbers are isomorphic as a ring to R[x]/(X^ + 1)
A vector space is just a ring homomorphism from a field into the endomorphism ring of an abelian group
A torus has zero global curvature
This is also true of spherical cows.Quote:
You can't comb the hair of a spherical dog without introducing a part
Another interesting fact about spherical cows is that you can split the cow into a finite number of pieces, reassemble it, and you'll have two new spherical cows of exactly the same volume as the first one.
This assumes that the cows are continuous.
And that's how math works. Isn't it pretty? You really can't disagree.Quote:
Originally Posted by khh
And that you want to work your way through all those homeomorphisms :PQuote:
Originally Posted by Xei
Actually I disagree with the axiom of associativity.
Prove it now, dick.
And udders. :VQuote:
And that you want to work your way through all those homeomorphisms
let's see, associativity....that's (a + b) + c = a + (b + c). It's gonna take me a second and I don't have my book that covers the fundamentals so I might not be able too. I'll do it in pieces.
we start with the symbols { and }. We introduce an operation, lets use S, which takes the given symbol, x, and returns {x} so that S(x) = {x}
We call S the increment operator and specify the rule that every { must be closed by a }
Fuck it. I haven't looked at that shit in ages. The problem is that to prove it (I'm sure you were being sarcastic) you have to construct the naturals from sets and that is frankly a pain in the ass that a lot of mathematicians couldn't do off the top of their head
I am a little embarrassed though....
I thought associativity is unprovable...
Perhaps not, dunno. Ask me in three years when I've got a title. :V
You studying maths Xei? Didn't know that.
It works like this. We collect a set of axioms and prove things assuming them. The axioms that describe the integers are collected together and called the ring axioms. The ones that describe the rationals and reals are called the field axioms. A set that satisfies them is called a ring or a field respectively. (only a ring doesn't necessarily need to be commutative in multiplication, only addition. If it is then it's specifically called a commutative ring)
Some other examples of rings are the ring of polynomials in the coefficient X under polynomial addition and polynomial multiplication, the ring of N x N matrices under matrix addition and multiplication, etc. If you want to apply the theorems that apply to a ring, field, group, etc., you have to prove that the set that you want to apply it to satisfies those axioms.
It is almost mandatory to assume that the number systems do satisfy them to learn basic math but it is necessary to construct them as I began to do above once you start to really get into it. You do it once in your undergraduate education (maybe) and then forget about it and take them as axioms ;) It's pretty dull and boring honestly.
A really good book is "Linear Algebra: An introduction to Abstract Mathematics" It doesn't prove that the axioms apply to the number systems but he does it for matrices ( and polynomials I believe) and it is a great introduction to the concepts. I think that you would get a lot out of it. He only assigns one computation in the whole book. All the rest of the problems are proofs.
If you want to see the numbers built up from scratch, check the book "Introduction to Algebra" by peter cameron. It has one chapter about it. The rest is algebra. He's at oxford and it's on oxford press. I'm not sure if it's still in print but you should be able to find a used copy. It's one of the best introductions to algebra in existence. I have no idea why it went out of print. If you can't find it, try "Undergraduate Algebra" by serge lang.
Algebra is my big thing. I love the stuff.
Thanks PS, I need to get some summer reading done actually. I prefer algebra to most other aspects of mathematics, although I understand it's a totally different beast in higher education. I think the thing I've enjoyed most so far was learning about the Maclaurin series and how they can prove euler's identity and then how you can use that for integrals and trigonometric sums and things.
Yeah and planning on mastering in neuroscience. Cambridge or Imperial College London, gonna be fun. :DQuote:
You studying maths Xei? Didn't know that.
Neuroscience, sweet. I'm gonna study medicine, specialize on psychiatry, then post-graduate on neuroscience. Maybe we'll meet one day ^^
Ok, all is well enough, but, help me out with this particular problem. Let's say that I have,Quote:
Originally Posted by PhilosopherStoned
1/(1-|x|)
I always imagine having different answers for x=1 and x=.999~
But the REASON I think that way is because we are only always able to portray the infinite string of 9's as a limited string (say ten, a hundred, or a thousand decimal places) in any equation, so I've had the bad habit of viewing .9999999999999999999999999999999999999999999999999 99999~ as a number that approaches 1 but is never at 1.
I can humbly admit that I was terribly, terribly wrong. But not without understandable cause!
Philosopherstoned already corrected me there, but there are no second place prizes. And I understand that the example isn't compatible with the problem. I'd imagine that one with a mind such as yours, Xei, would opt towards helping me without the negative attitude.Quote:
Except parabolas don't even have any asymptotes.
*gets popcorn and scratches head with confused look*
Taking it like a man ;) It's a very understandable cause by the way. Math is hard enough without the fairly bad job they do of teaching it in school making the situation worse. Our brain isn't designed to handle it. You really do have to learn to just drop your intuition and go with the axioms, theorems and definitions. Trying to understand why your intuition is wrong, when it is, is also great for developing it. Over time, it becomes a friend, not a foe. The fact that you and khh were able to understand the proofs is fairly huge. Most people wouldn't be open minded enough, honestly.Quote:
Originally Posted by Invader
I might jump off a cliff after I open up this can of worms with you again, but how does the number get (not that it is moving) infinitely close to 1? Apparently it does, but how? Infinity goes on forever. At any 9, you can move to the next 9, and 500 quadrillion googol octillion 9's later. The 9's extend forever. It has no end, so how is it 1? It just sounds so much like talking about the end of space. 1 is a very exact number.Quote:
Originally Posted by Xei
0.999... is also an 'exact' number (being the same number, 1).
Quote:
how?
That's the answer really. Each consequtive 9 decreases the difference between the number and 1 by a factor of 10. You do this infinite times and you've decreased the difference by a factor of infinity; i.e. there is 0 difference.Quote:
Infinity goes on forever.
I think you're asking at what point leading up to infinity is it 0; there is no point. That's the whole point of infinity, it isn't a number as such. It just represents what would happen if you did it again and again and again forever. You can't in practice, but it's just a concept.
The above is all intuitive however. The real answer to your question is simply the mathematical proofs which have been offered. Mathematics is abstract. Think what would happen if you tried to discuss the Banach–Tarski paradox in this way; it doesn't work.
I sort of see what you are saying, but it's that concept of infinity that is so baffling. Like, if you traveled an infinite distance in one direction, you would go through all of space in that direction. That makes sense in a way, and theoretically it is true, but it is still a big WTF since space does not end.
I remember proving this in Algebra 2. Fun.
inb4lock
Hah, I'm trying to be. I can't say I'm not a little worried, though. I'm supposedQuote:
Originally Posted by PhilosopherStoned
to be going into calculus II after not having had any involvement with serious
mathematics for the past two years. I guess I know who to go to if I need that
little bit of help now, aye? :chuckle:
I went back and edited my proof. It's much clearer now, I think. I had been up for about 36 hours when I first wrote it. I think it's the clearest proof of it on this thread because it's a 'why' proof that actually explains why it is the way it is as opposed to relying on calculus like the OP's proof, a formula like Kromoh's proof or algebraic trickery like xei's proof. xei's proof is the most elegant though without a doubt. My proof is essentially along the same lines as xei's informal argument that almost has you convinced. Making it formal might help. Unfortunately, it's the longest proof on the thread, but that's almost always the way it is with explanatory proofs vs. clever proofs.Quote:
Originally Posted by Universal Mind
I'll happy to help and I'm sure a lot of other people will as well. It's funny about calc II because I made a B in that class. First, I didn't do my homework which counted for 5% of the grade. Then, I didn't show up for class and that counted for like 2% or something. Then I missed the date on a test and got a zero for it. He dropped the lowest test grade so I was going to be fine. Then I forgot some formulae for some curves on the analytic geometry test because I was stoned and hadn't slept much. I was able to derive some of them in the allotted time but not enough to get above a 70%. Averaged in with my other perfect scores, I made a high B.Quote:
Originally Posted by Invader
I should point out that that was not the first time that I had studied that material and was already studying toplogy and algebra in my free time so it was, at that time, pretty elementary stuff for me. I had to work hard to understand the material the first time around.
How does it work in the U.S. anyway?
I sometimes get the feeling that we do less in Britain. Most people who do any proper maths do Maths A-level which is a qualification which takes up two years (you start when you're 17), but I did another one callled Further Maths... it basically takes you up to euler's identity.
But I talked to somebody from the U.S. at my Cambridge interview and he'd done stuff like harmonic series and convergence which we hadn't even touched upon.
I'm not the best person to ask. I dropped out of highschool as soon as I was legally allowed to and then tested into calculs I (They wouldn't let anybody test any higher than that) when I started community college a few years later.
I know that the average "good" student starts with calculus one when the first get to college. We cover convergence of series and sequences in calculus II. We don't do partial derivatives until calculus III. We do the jacobian in calculus three, but only the determinant of it for change of variables in multiple integrals. We cover lagrange multipliers and a week form of stokes theorem that only relates the interior of subsets of R^3 to it's boundary.
After calculus II you can take linear algebra and differential equations. I think the only reason they make you wait till after calculus II for linear algebra is to weed people out. Logically speaking, I think linear algebra should come first because the language that you learn in it does a lot of good for calculus. If I had my way, we would skip single variable calculus all together and then go straight to multivariable calculus after linear algebra.
What age is this..?
I think 'college' means quite different things to us. In the UK it means a couple of years when you're 16 to 18 getting qualifications, usually to go onto university.
College is 18-22 generally. 16-18 is high school.
when you studied calculus, did you do the epsilon-delta formulation of limits, or did they just wave their hands over them? It was just hand-waving in the US
I'm 18 so basically I've just finished what you would call High School. University starts in October.
We've hardly done any limits at all. I can tell you what GPs converge to, I can tell you that harmonic series don't converge, and I can do some trigonometric series using complex numbers. Think that's it.
Oh, I'm already convinced. I just still see paradox in it. I don't question the truth of it any more.Quote:
Originally Posted by PhilosopherStoned
Neuroscience sucks. Xei do pure maths not applied maths.
I wouldn't say Further maths takes you up to euler identity. As hyperbolic functions is more advanced than euler identitiy. Sadly, I got a handout that had some basic complex analysis i.e. how to use complex numbers to integrate functions.
Xei results day is in like 14 days and some hours. I'm really nerous but I really shouldn't because I'm pretty sure I will get AAA. Unless, something crazy happens. Good luck Xei.
Yeah we had to do complex calculus although normally it was quicker to do it the simpler way. tbh hyperbolics in FM are very basic, it's essentially just the trig identities all over again...
Good luck to you too. I think I got 1,2 or lower in STEP (missed my offer) and AAAAA or AAAAB with the B in Biology. I'd say it's a toss up now if I get into ICL or Cambridge.