Do this maths. It R Hard.
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Factorization in the general case is trial and error. Remembering the formula for solving quadratic equations is meh.
QFT
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So just add one step:
"We've established the answer is 10 or -9. For both, n^2 - n - 90 = 0 holds. QED."
It's like saying "you can't prove 4 is an even number by proving all numbers of form 2n are even, then proving 4 is 22, you need to prove 4 and 4 only." TDEMSYR.
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So just add one step:
"We've established the answer is 10 or -9. For both, n^2 - n - 90 = 0 holds. QED."
It's like saying "you can't prove 4 is an even number by proving all numbers of form 2n are even, then proving 4 is 22, you need to prove 4 and 4 only." TDEMSYR.
It's not like that at all. And your one step is even more wrong and tautological than what Urchin was saying.
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Can we get a recap post with all the logic now? I got lost somewhere.
OK. I'm even going to number these properly.- Assume n² - n - 90 = 0.
- There are two solutions to the equation, -9 and 10.
- We're talking about sweets; -9 sweets is nonsense.
- We're left with 10 as the only sensible value for n.
- Since there are 10 sweets, and 6 are orange, 4 must be yellow. Or blue. Or pink. Or dayglo black. Doesn't matter; they're not orange.
- The probability of the first sweet taken being orange is therefore 6-in-10.
- The probability of the second sweet taken being orange is therefore 5-in-9.
- Multiply those probabilities to get the probability of two orange in a row: 30-in-90.
- Simplify to 1-in-3.
- We now know that the only sensible solution to the equation supports the probability stated.
- Can
nbe anything else? No. Why? - The number of orange sweets is fixed by definition.
- The number of non-orange sweets is therefore the only variable.
- An increase in non-orange sweets lowers the probabilities of an orange one being selected.
- A decrease in non-orange sweets raises the probabilities of an orange one being selected.
- The relationship between the probability and the number of non-orange sweets is polynomial.
If that is not a complete logical deduction that the only possible value for
nis 10, then I'm on Europa.When you have eliminated the impossible, whatever's left must be the truth.
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f(x) = 1/3rdy(x) = 1/3rd
how do you prove which is the correct equation?
they both are.
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There are two solutions to the equation, -9 and 10.
If that is not a complete logical deduction
Well it's not, you started with the answer without actually deducing it from the equation.
You need:
-1. Given n*n - n - 90 = 0
0. Via the quadratic formula, n is either 10 or -9And most professors would want you to show your work on step 0 as well. "@accalia said it's x" isn't actually a logical deduction step :)
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Factorization in the general case is trial and error
Yes. It helped that I already knew the answers I was looking for, and that the n^2 didn't have a multiplier
Is your method completing the square? It's a long time since I've done that, and I'm not confident I could do it as fast as using the quadratic formula without practice
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you started with the answer without actually deducing it from the equation
The very definition of begging the question
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What's this bit about:
@RaceProUK said:The relationship between the probability and the number of non-orange sweets is polynomial.
How do you work that out and why does it matter?
Polynomial's got nothing to do with it. What matters is showing that it's bijective. Polynomials are not all bijective (indeed, the quadratic that is the solution is not, without additionally restricting the domain to positive numbers) and not all bijective functions are polynomial.
If that is not a complete logical deduction that the only possible value for n is 10, then I'm on Europa.
It's a hideous one that relies on the probability function being intuitive. As a pure mathematician I find it deeply disgusting. But it is, kinda-sorta-just about, a solution to the problem.
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And most professors would want you to show your work on step 0 as well. "@accalia said it's x" isn't actually a logical deduction step
So I worked from what someone else had worked out. Doesn't make my reasoning any less valid, since what I built on is independently verifiable.
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Is your method completing the square? It's a long time since I've done that, and I'm not confident I could do it as fast as using the quadratic formula without practice
No clue what it's called in English, sorry. I'm not even sure I recall what it's called in Dutch.
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@Jaloopa said:
Is your method completing the square? It's a long time since I've done that, and I'm not confident I could do it as fast as using the quadratic formula without practice
No clue what it's called in English, sorry. I'm not even sure I recall what it's called in Dutch.
Yes, that's completing the square.
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. Doesn't make my reasoning any less valid
It means your answer is incomplete. Which I think is what people are trying to say: you didn't begin at the beginning of your solution, you began partway in and assumed the first bit was true. Which is an awful way to communicate a solution.
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How do you work that out and why does it matter?
It's an inverse relationship, and one that is obvious given the limitations given by the original question. And it matters because it shows that the relationship is a simple one without any fancy loops to mess things up.
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So basically I've spent all afternoon being told I'm wrong because people couldn't be bothered to read my posts in the correct context.
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Did that briefly at A-Level, decided it wasn't the best way to solve quadratics for my way of thinking, and haven't done it or seen it in over 10 years. I'm quite impressed I recognised it at all
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So basically I've spent all afternoon being told I'm wrong because
people couldn't be bothered to read my posts in the correct contextI'm wrong.<flamebait
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I have spent all fucking afternoon showing what I have deduced is correct. I have plugged every fucking gap. I have excluded every fucking possibility except one. And yet no-one will fucking accept it.
Fuck this, I'm done. I'm muting this before I really get pissed off.
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Just toying with this at the moment - no idea if it's in any fit state to put on here yet:
http://pjh.homeip.net/t/mathjax-testing/76/3?u=pjh
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So basically I've spent all afternoon being told I'm wrong because people couldn't be bothered to read my posts in the correct context.
No.
Accalia's solution was wrong too.
Your original addition to it was utterly pointless, a tautology that added nothing to the wrong solution. It was just as wrong, and I think you thought you were proving the forwards implication, which makes it even more wrong.
It was only a gazilion posts down the line that you added the important part that made your answer right.The original answer was wrong, it was not misunderstood by anyone but yourself.
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Just toying with this at the moment - no idea if it's in any fit state to put on here yet:
wheee! i've been waiting for that for ages! :-D
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Accalia's solution was wrong too.
It was incomplete. I completed it.
@CarrieVS said:It was only a gazilion posts down the line that you added the important part that made your answer right.
Yes, because evidently I had to explicitly state the 58000 truths that are obvious.
And people are still calling it wrong. That's why I'm not doing this anymore.
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You were still begging the question. Everything else you did wrong came from the initial incorrect assumption that (a=>b) == (b =>a)
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When you have eliminated the impossible, whatever's left must be the truth.
I eliminated all but one possible solution. So I got there the long way. I don't fucking care. Was my answer right? Yes. Are my deductions consistent and logical? Yes.I answered the fucking question. Deal with it.
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I answered the wrong fucking question. Deal with it.
You showed that if the equation is correct, the original assumptions are valid. That is not the same as showing that, given the original assumptions, the equation is valid.
Eventually, you did plug the gaps, but you still won't accept that you were originally wrong. Therefore you get angry when people keep trying to point that out
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It was incomplete. I completed it.
Your original answer was no more complete. It added nothing. It was nothing but a tautology, and if you still insist it actually added anything to the answer, that only shows that you still don't understand it and it's only coincidence that something you eventually wrote is, just about, a correct solution.Yes, because evidently I had to explicitly state the 58000 truths that are obvious.
The fact that it's a biimplication is non-trivial and that's how maths works.And people are still calling it wrong.
Who? I'm calling it disgusting and much less clear than the simple constructive proof, but your eventual answer was correct, though of debatable rigour.I eliminated all but one possible solution. So I got there the long way. I don't fucking care. Was my answer right? Yes. Are my deductions consistent and logical? Yes.
I answered the fucking question. Deal with it.
None of this was the case until you added the 'proof' that it was a biimplication.
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You know how we, as programmers, get annoyed with the "it compiles, ship it" mentality of the kind of hacks who make it to the front page? To mathematicians, you're doing the same thing. Maths is all about deducing, logically and clearly, that assumptions lead to results. If something seems obvious, it still needs proving and is often harder to prove than something that isn't as intuitive.
You're saying it's not a bug until it's uncovered. We're saying you're not complete and therefore don't have a correct solution
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[spoiler]
probability p1 is 6 / n
probability p2 is p1 * (5 / (n-1)
p1p2 = (6 / n) * (5 / (n - 1))
p1p2 = 1 / 3
1 / 3 = (6 / n) * 5 / (n - 1))
1/3 = 30 / n * (n - 1)
1/3 = 30 / n^2 - n
1 = 90 / N^2 - n
N^2 - n = 90
N^2 - n - 90 = 0
[/spoiler]There's a flaw in your work.
[spoiler]At the 8th line, n^2 becomes N^2. According to standard mathematical notation, n and N should represent unique values. This means that in the 8th line you mysteriously changed one unknown into another unkown. However, since you left the other n value untouched, you now have two unknown values in your simplified formula. Your end result not only doesn't match the formula specified in the question, but it is unsolvable on its own.[/spoiler]
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still alloys me
You're being alloyed? I'm not sure how that works with a cake. Or a fox.
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You're being alloyed? I'm not sure how that works with a cake. Or a fox.
May I remind the Hatter of one of said fox's previous avatars?
Also, how did I miss that one?
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@Maciejasjmj said:
So just add one step:
"We've established the answer is 10 or -9. For both, n^2 - n - 90 = 0 holds. QED."
It's like saying "you can't prove 4 is an even number by proving all numbers of form 2n are even, then proving 4 is 22, you need to prove 4 and 4 only." TDEMSYR.
It's not like that at all. And your one step is even more wrong and tautological than what Urchin was saying.
- You solve the original problem with some provably correct method. You obtain a set of values for n.
- You enumerate the set of possible values and show for each one that n^2 - n - 90 = 0 holds.
- You've established that for each possible solution, n^2 - n - 90 = 0 holds.
By doing this, you reduce the problem to "prove your way of solving the original problem is correct". Depending on the method used, this may be as simple as referring to prior work.
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You assume the equation held, work out the values of n that would give you that probability if the equation held. The you take those values and show that if the equation holds, then the equation holds.
Which you could have said without doing any working out at all, because it's a tautology, and pointless.You didn't prove that the value of n that gives you the correct probability if the equation holds is the only value of n that makes the probability correct, so you haven't proved that the equation holds.
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What's the question again.
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What is the answer to Life, The Universe, and Everything? Show your work.
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$42_{10} = 111111111111111111111111111111111111111111_1$
Pf/
join . (fmap show) . take 42 . repeat $ 1
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you call that a proof?
you call that a proof?!
..... wait.... What?
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It's not just a proof, it's intuitionistically valid. I didn't even have to resort to LEM.
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@RaceProUK said:
When you have eliminated the impossible, whatever's left must be the truth.
I eliminated all but one possible solution. So I got there the long way. I don't fucking care. Was my answer right? Yes. Are my deductions consistent and logical? Yes.I answered the fucking question. Deal with it.
Here's the original problem:
This is how I (and others) read it:
**Given:** - Bag contains n sweets - 6 sweets are orange - (n - 6) sweets are yellow - The probability of drawing one orange sweet and then drawing another orange sweet is 1/3
Problem:
Find the formula to solve for n (n^2 - n - 90 = 0)This is what @accalia seems to have seen:
**Given:** - Bag contains n sweets - 6 sweets are orange - (n - 6) sweets are yellow - The probability of drawing one orange sweet and then drawing another orange sweet is 1/3 - `n^2 - n - 90 = 0`
Problem:
?????????This is the problem you solved:
**Given:** - Bag contains n sweets - 6 sweets are orange - (n - 6) sweets are yellow - The probability of drawing one orange sweet and then drawing another orange sweet is 1/3
Problem:
Show thatn^2 - n - 90 = 0provides a value that meets the 1/3 probability.No one is arguing that your logic or math is incorrect. Everyone is arguing that you answered the wrong question.
Looking at the original image, you can see there is the setup for the question, and then the question itself is offset by (a). This indicates that there are likely additional question relating to the word problem, such as "How many sweets were in the bag?" With that in mind, it's easy to deduce that the solution you provided isn't the one that the question is asking for.
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What is the answer to Life, The Universe, and Everything? Show your work.
WRONG. Your question should be:
What is the answer to The Ultimate Question of Life, the Universe, and Everything? Show your work.
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Looking at the original image, you can see there is the setup for the question, and then the question itself is offset by (a). This indicates that there are likely additional question relating to the word problem, such as "How many sweets were in the bag?" With that in mind, it's easy to deduce that the solution you provided isn't the one that the question is asking for.
And had those extra questions been shown, I may have approached the first one differently.I answered the question I had in front of me; if there is information missing, then how can I be expected to get the correct interpretation?
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No one is arguing that your logic or math is incorrect.
Objection!
They keep claiming that they have proved that the equation holds, which is utterly illogical because they started with the equation as a given. You cannot prove a thing if your logic requires assuming it's true. (You can prove a thing by assuming it's not true, but that's entirely different from what anyone tried to do here.)
had those extra questions been shown, I may have approached the first one differently.
I answered the question I had in front of me; if there is information missing, then how can I be expected to get the correct interpretation?
No you didn't, and there was nothing missing.
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nah. that's the question for the final, not the midternm. ;-)
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\newcommand{\P}{\mathbf{P}\left[#1\right]}
Anyway, it's trivial.
Hanna samples without replacement. Letting $O_i$ be the event that Hannah's $i$-th sample is orange, we have
\begin{align*}
\P(O_1 O_2) &= \frac{1}{3} \
&= \frac{6}{n}\frac{5}{n-1} \
30 &= \frac{1}{3} n(n-1) \
90 &= n^2 - n \
0 &= n^2 - n - 90
\end{align*}I got my pencil and paper out for this?
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Apart from questions (b) and later.
But you weren't trying to answer them, you were trying to answer question (a). There is no information missing in respect of question (a).
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And had (b) and onwards been included, I'd have known what else they were looking for, and would have had more information to select a suitable approach for (a).
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Going by this post, the proof is fine.
Essentially, what the problem asks is "Prove that for every value of n that results in a probability of 1/3 for event E, the equation holds". The easy way to do it is to derive one equation from the other:
prob(o0 o1) = 1/3 = p(o0) * p(o1)
p(o0) = 6 / n
p(o1) = 5 / (n-1)(6 / n) * (5 / (n-1)) = 1/3
(65) / (n(n-1)) = 1/3
30 / (n^2 - n) = 1/3
30 / (1/3) = n^2 - n
30 * 3 = (n^2 - n)
90 = n^2 - n
n^2 - n - 90 = 0The other way to do it (what they did) is to find every value of n for which the equation is true, every value of n for which the probability of E is 1/3, and show that the solutions to the equation is a superset of the solutions for the probability. In the post I link above, they do this. They present an argument for why the probability is only true with one n (at the end), and find that the equation holds for that single value. Proof by exhaustion.
It's just a lot more work and more error prone.
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Finally! Someone who actually understands what I was trying to do!
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Wait, on what basis is that proof assuming that p(o0 o1) = p(o0) * p(o1)? That's not true in general.
