Do this maths. It R Hard.
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If I'm teaching my brother how to make a two-column site, and he installs Wordpress and finds a two-column theme, he's clever, but he hasn't shown me that he learned what I tried to teach him, has he?
That's a very bad analogy. If he installs a pre-made solution, he hasn't learned to make a two-column site. If instead he made two columns from scratch but did it in a different way to you, he has.
The amount of time I wasted in school mucking about with the Belgium quadratic formula or working out if a quadratic would factorise because they wanted me to use a specific method instead of completing the square like a sane person.
That's just one example. It's a crying shame the way they teach what passes for maths in schools. Memorising a formula is not maths. Maths is finding the formula. And if two people each find a slightly different version of it, they're both equally right.
the best maths professor i had in school focused on teaching how to think through the problems... he taught me how to reason through the problem to arrive at the correct solution, he did not teach me 30 formulas that i knew how to apply be rote and i could use to instantly solve any problem that exactly fit the form that the formula needed.
Sound almost like you had a real mathematician teaching you.
How do you test someone's ability to think through a problem if your test grading policy is "If you get the right answer, it's right, no matter what you did in the process"?
You can still require people to show their working if any correct process works.
And I'm willing to bet my logical deduction method would be marked down in favour of @algorythmics and @CarrieVS algebraic approach, which is unfair, as both methods are equally valid in their conclusions.
If there are 10 sweets, then the probability of picking two orange sweets in a row is 6-in-10 then 5-in-9; multiply to 30-in-90, then simplify to 1-in-3.
This?
That's not equally valid. It could be if you did a bit more work but without proof that 10 is the only possible number of sweets in the bag it does not answer the question. And proving that would probably take you through an algebraic solution anyway. (You can't prove it by solving the equation, because then you're using circular logic)
Besides that you must have got 10 either through trial and error or by working backwards from the answer and then forwards again, either of which is more effort and less elegant than simple construction.
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That's not equally valid. It could be if you did a bit more work but without proof that 10 is the only possible number of sweets in the bag it does not answer the question. And proving that would probably take you through an algebraic solution anyway. (You can't prove it by solving the equation, because then you're using circular logic)
It was a follow-on from @accalia solving the equation forn; since the two solutions are 10 and -9, and the latter of which is ludicrous, 10 is the only possible solution to the equation. Combine that with my logical deduction, and you have sufficient proof that n² - n - 90 = 0 and the probability of two orange sweets in a row is 1-in-3 are self-consistent and non-contradictory.Is it really so bad I solved an algebraic problem via logical deduction?
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It was a follow-on from @accalia solving the equation for n; since the two solutions are 10 and -9, the latter of which is ludicrous, then 10 is the only possible solution to the equation.
That's circular logic.
You're using the fact that the equation holds to prove that the equation holds.
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So it's proof by tautology. It's still proof.
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You're answering a question they didn't ask.
they might not have asked it, but it feels to me like that was what they're asking.
and it's not helped by my extremely mild case of dyslexia (with my spellaring, is that any surprise?)
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So it's proof by tautology. It's still proof.
No it's not.
You're basically saying 'if this equation holds, the equation holds'. That doesn't prove anything.
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[spoiler]
Chance to grab orange the first time 6 / n = c1 Chance to grab orange the second time 5 / (n - 1) = c2 c1 * c2 = 1/3 6/n * 5/(n-1) = 1/3 30/n(n-1) = 1/3 90 = n*n -n[/spoiler]
That took me embarrasingly long.
Filed under: Spoiler tags don't work, but I don't care
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they might not have asked it, but it feels to me like that was what they're asking.
Stop feeling and start reading?
and it's not helped by my extremely mild case of dyslexia (with my spellaring, is that any surprise?)
Maybe, but this is a big difference.
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That's circular logic.You're using the fact that the equation holds to prove that the equation holds.
not quite though. it's not a proof that the equation is the only solution, but it's certainly proof that the equation is a solution.
":If i accept the equation as true then the probabilities work out, and if i accept the probabilities as true the equation balances. this the equation must be a solution to the problem"
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You're basically saying 'if this equation holds, the equation holds'.
No, I'm sayingIf the equation holds, the probability of two orange in a row is 1-in-3.
The question states that that probability is correct; what am I missing, other than pedantry about the exact method?
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Carrie's point is you haven't shown that the equation holds, which is what the answer asks for.
It doesn't say "show n = 10" or "show the probability of getting 2 orange sweets is 1/3", which are the 2 things your answer actually shows. It asks you to show that the probability can successfully be manipulated to that formula.
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Fine. Let's just throw out a perfectly valid reasoning because of some pedantic dickweedery.
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No one's throwing out the reasoning. You just didn't answer the question that was asked.
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@CarrieVS said:
You're basically saying 'if this equation holds, the equation holds'.
No, I'm sayingIf the equation holds, the probability of two orange in a row is 1-in-3.
That's not the question asked! The question asks to show 'if A then B', and you showed 'if B then A', and neither of those automatically implies the other.
Fine. Let's just throw out a perfectly valid reasoning because of some pedantic dickweedery.
It's not pedantic dickweedery, your answer is wrong.
That is, claiming it proves that the equation holds is wrong. There's nothing wrong with the maths in it, but it doesn't show what you're claiming it does, therefore you're wrong.
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"Can this building support a roof weighing 2 tonnes?"
"If this building can support a roof weighing 2 tonnes, it can have a cross sectional area of 325m2, if the roof is made out of clay roofing tiles, it will weigh 2 tonnes. Yes, this building can support a roof weighing 2 tonnes."
It doesn't actually answer the question of whether the building supports the roof. It works backwards from an assumption the roof is supported and finds no errors, but that's just the no errors falacy. Finding no bugs does not mean no bugs exist.
The important thing is, even though in this case there is only 1 possible answer, the process falls down, because that isn't always so.
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The question asks to show 'if A then B', and you showed 'if B then A', and neither of those automatically implies the other.
Ifnwas anything other than 10, then the formula and the probability would be contradictory; in this situation, I recognised that the relation is transitive.
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"Can this building support a roof weighing 2 tonnes?"
Is an engineering question, not an algebraic/logical one; I'd approach it as an engineer, not a logician.
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But you can't state that the equation holds for any value of n that gives that probability until you've proved it.
It happens the solution of the equation is the only such value buy you don't know that unless you prove that the equation holds without depending on the assumption that it does.
[edit: thought I was making a new post and I'd been editng this one.]
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If n was anything other than 10, then the formula and the probability would be contradictory; in this situation, I recognised that the relation is transitive.
So you're relying on test grader telepathy?
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There are only two solutions for the equation, -9 and 10. Since you can't have -9 sweets, you must have 10. Since you have now reduced the number of possible values to a single number, the only thing left is to show you get the probability of 1-in-3, which I did.
When you have eliminated the impossible, whatever's left must be the truth.
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but the only way to find the results of the equation is having the equation. How do you know the equation is correct?
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If the equation was wrong, then I'd get a different probability than 1-in-3.
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You like finding creative ways to fail, don't you?
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f(x) = 1/3rd
y(x) = 1/3rdhow do you prove which is the correct equation?
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How the fuck is that related to the original problem?
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If the equation was wrong, then I'd get a different probability than 1-in-3.
Not necessarily.
What you've done is prove a single 'if', a unidirectional implication and assume that it must be a bi-implication - if and only if . You can't do that, it's incorrect.
'If x = 1, then x^2 = 1' is true, but 'if x^2 = 1, then x = 1' is not. You can never assume a bi-implication just by proving one direction.
In this case it is a bi-implication, but you didn't show the other direction and it's non-trivial that it is.
At the risk of appealing to authority, I have a certificate saying that I'm officially a Master of this subject. I am not wrong about the difference between if and iff.
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f(x) is 30 / n2 - n
n = x
y(x) is another equation with the same result for n = 10. and n = 20, and n = 23459.
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The difference is the same as the difference between a heuristic and a solution.
Your answer is a heuristic which is true in this case, but applying the same logic in all similar circumstances will not hold true.
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y(x) is another equation with the same result for n = 10. and n = 20, and n = 23459.
And? The question doesn't say anything about other formulae also yielding the correct result.I showed that, given the equation and the probability, they are self-consistent. At no point have I excluded other equations from matching.
At this point, I can't help but wonder if you're doing this not because you think my approach is wrong, but simply to see how far you can push me before I snap. And I don't appreciate that.
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At this point, I can't help but wonder if you're doing this not because you think my approach is wrong, but simply to see how far you can push me before I snap. And I don't appreciate that.
I promise you, that's not the case. You are not correct. Pinky swear.
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I've yet to see where I'm wrong.
Two of you used the probability to derive the formula.
I used the formula to derive the probability.
Both methods yielded the same answer.So why are you right and I'm wrong?
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Two of you used the probability to derive the formula.I used the formula to derive the probability.Both methods yielded the same answer.
So why are you right and I'm wrong?
Because the question was to show that the probability implies the formula, which is what we proved.
That is not the same thing as the formula implying the probability, which is what you proved.
You proved that the only (sensible) solution to the equation is a value of n that gives you the right probability.
You didn't prove that the only value of n that gives you the right probability is a solution to the equation.
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I map from n² - n - 90 = 0 to the probability, and get the same result
You're begging the question.
When I had problems like this at A level, I often used to put the answer at the bottom, the starting data at the top and work from the bottom up and top down until I met somewhere in the middle. As long as it's in the right order on the paper, it's all fine
At this point, I can't help but wonder if you're doing this not because you think my approach is wrong, but simply to see how far you can push me before I snap. And I don't appreciate that
It's not all about you. People are disagreeing because they're right and you're wrong
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There are a fixed number of orange sweets, so varying
ncan only affect the number of yellow sweets (or whatever colour, I can't be arsed to scroll up and find out). Add more yellow sweets, the probability of two oranges in a row drops. Reduce the number of yellow sweets, the probability rises. It's a simple relationship.Now combine that, which is frankly so obvious it shouldn't even need saying, with the rest of my working, and the conclusion is obvious. And it's what I've been saying all along.
n === 10is the only possible solution.
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At this point, I can't help but wonder if you're doing this not because you think my approach is wrong, but simply to see how far you can push me before I snap. And I don't appreciate that.
not what I'm trying to do, honest. let me see if I can bring in an example from the other problem.
that first step is flawed. You correctly ruled out some dates, but you ruled them out for the wrong reasons, and missed other invalid dates as a result, this made you think that some dates were valid when they weren't, and led to faulty reasoning later on.
It's a similar circumstance, in the original problem people ruled things out but not thoroughly enough, which allowed them to come to a single answer but that answer was incorrect.
In this case, doing essentially the same thing (by solving the equation and using the answer to verify the equation) you are ruling out all other possible equations that could exist with no mathematical basis.
There is no mathematical basis for eliminating the value -9 as a possible answer, there is only a physical one.
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There is no mathematical basis for eliminating the value -9 as a possible answer, there is only a physical one.
True, but the question is about sweets; the physical component is important.
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[spoiler]
The chance of the first sweet being orange is6/n. The chance of the second sweet being orange, given that the first is orange, is5/(n-1). The chance of both being orange is thus5/n * 6/(n-1), which is also given to be1/3. Working from there:
[/spoiler]
[spoiler]
5/n * 6/(n-1) = 1/3
(5*6)/(n*(n-1)) = 1/3
30/(n²-n) = 1/3
90=n²-n
0=n²-n-90
[/spoiler]
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There are a fixed number of orange sweets, so varying n can only affect the number of yellow sweets (or whatever colour, I can't be arsed to scroll up and find out). Add more yellow sweets, the probability of two oranges in a row drops. Reduce the number of yellow sweets, the probability rises. It's a simple relationship.
Now combine that, which is frankly so obvious it shouldn't even need saying, with the rest of my working, and the conclusion is obvious. And it's what I've been saying all along.
I'm not sure that's rigorous, but ok, I guess.
But it does need stating. It's non-trivial that the probability as a function of n is 1-1, but if you prove that it is then a value of n that gives the correct probability must be the only one.But that's a really convoluted and inelegant solution. Maths is just as correct if it's not elegant, but it makes my skin crawl, given there's a constructive proof that's as simple as anything and much clearer, and if I was grading a test with that answer I'd wish I could mark it wrong.
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n === 10 is the only possible solution.
This is true. Your reasoning to get to that is faulty though
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Where? Over about a gazillion posts now, I've closed what few holes were in my logic; there is simply no other conclusion.
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Over about a gazillion posts now
As opposed to the <10 lines everyone else has used. If you'd put your original answer in the test you would have got it wrong, since the marker can't come back to you and say "what about this? Did you mean that?"
I'd still argue that although you've proved the number of sweets you still haven't demonstrated the thing you were actually required to answer, so even with the clarifications you wouldn't get full marks
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I demonstrated it by eliminating every possible alternative
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Can we get a recap post with all the logic now? I got lost somewhere.
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Logic is like a city with a one-way system. You can't assume you can get back from somewhere the same route came. You have to prove that none of the streets are one-way. It's not a question of a hole in your logic, everything you'd proved until that last post did nothing to show that you could get from the probability to the equation.
And I'm still not convinced the last post is rigorous, or that you really understand why you were wrong. Probability is frequently unintuitive, and a lot of non-mathematicians seem to have difficulty with the directionality of implications.
I'd still argue that although you've proved the number of sweets you still haven't demonstrated the thing you were actually required to answer, so even with the clarifications you wouldn't get full marks
She's proved the reverse implication, and kinda-sorta-explained why it must be a bi-implication. It's a convoluted answer and ugly as sin and I'm not convinced about the explanation being sufficient proof, but she's pretty much shown it.
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Proceeding to solve for n, because I can:
n²-n-90=0
n²-n=90
(n-½)²-¼=360/4
(n-½)²=361/4
n-½=±√(361/4)
n=½±19/2
n=10 or n=-9
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Logic is like a city with a one-way system. You can't assume you can get back from somewhere the same route came
In mathematical terms:
(a => b) != (b => a)
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Proceeding to solve for n, because I can:
n²-n-90=0
n²-n=90
(n-½)²-¼=360/4
(n-½)²=361/4
n-½=±√(361/4)
n=½±19/2
n=10 or n=-9n2 - n - 90 = 0
(n-10)(n+9) = 0
n = 10 or -9I like factorisation
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n²-n-90=0
n²-n=90
(n-½)²-¼=360/4
(n-½)²=361/4
n-½=±√(361/4)
n=½±19/2
n=10 or n=-9Half credit. You didn't spot the implicit constraint that it must be a positive number.
In mathematical terms:
(a => b) != (b => a)</blockquote>Yes. I was trying to explain precisely that by way of an analogy. Single-direction implication seems to really confuse a lot of non-mathematicians.
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Factorization in the general case is trial and error. Remembering the formula for solving quadratic equations is meh. I started remembering this approach when I needed to solve quadratic inequalities a couple of times in a row.
