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    Operations over infinite decimals numbers

    0.999... does equal 1, see [Broken]
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    Teaching myself calculus?

    Calculus by Michael Spivak is really, really good if you want a thorough understanding of the concepts. The problems are all useful and challenging, and you learn a lot more than calculus in the process. However, I've heard it's quite hard compared to other calculus books, and you won't like it...
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    Absolute value

    With absolute values, only the magnitude of the change matters. It doesn't matter if the change is positive or negative. So H must change more than TS.
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    Prove that a rational number squared has each of its prime factors with even exponent

    So how does this work with \sqrt{18}, \sqrt[3]{7}, or \pi? I'm just trying to show that you may be overlooking some possibilities.
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    Alternative for sphere volume:FAIL

    This is pretty much integral calculus, summing an infinite number of infinitely small areas. You should learn it, it's really useful.
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    Consecutive integers divisible by a set of Primes

    I'll try to come up with a better one, mine's terrible. How on Earth did you work out frg(15)?
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    Consecutive integers divisible by a set of Primes

    I calculated the first 8 and put them in to OEIS, and got: What you're after is not the gaps, but the difference, so it's one more than the terms in the sequence I linked to. I'm not sure about an efficient algorithm, my jumbled together program could only do 8 before taking...
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    Alternative for sphere volume:FAIL

    So it seems to be (area of circle) * (circumference) * (1/2). It's less a question of why it doesn't work, than why your brother thought it would work. Maybe if you posted the derivation, we could point out the problem with it.
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    This bigger than grahams number?

    Stop it. You're absolutely unimaginably nowhere near Graham's number, and it's pointless to try and come up with a larger number.
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    Possible values of X and Y for the problem

    99 = 3 * 33, so 1 and 11 aren't the only values x or y can take. There's 1 solution to your problem. Don't be surprised that you're getting "ad hoc" methods, you've thrown together a few random arbitrary conditions, especially the "twin number" bit.
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    Possible values of X and Y for the problem

    If x and y are whole numbers, then x/y can't be irrational.
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    Sum of Sums over Primes that Divide the Index

    I think that's equivalent to \sum_{p=2}^{n} \frac{\left \lfloor n/p \right \rfloor}{p} , where the square brackets represent the floor function, and p runs through the primes less than or equal to n. I don't know if that helps at all, and no doubt it can be simplified more so.
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    Logic behind the number of combinations of numbers

    It might help to imagine a tree diagram, with all the possibilities the numbers could be.
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    Has any equation ever been proved

    Now I hate this kind of discussion, so I won't really get involved, but mathematics does not attempt to prove anything "in terms of the physical world", it has no concern for silly things like atoms and quantum systems.
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    Maths proof for fractions

    My LaTeX always looks so ugly. n = \frac{(\frac{a}{b})}{(\frac{c}{d})} n\left(\frac{c}{d}\right) = \frac{a}{b} nc = \frac{ad}{b} n = \frac{ad}{bc} = \left(\frac{a}{b}\right) \times \left(\frac{d}{c}\right)
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    Simplifying equations. Orders of operation.

    The generally accepted order of operations are BIDMAS (or BODMAS, same thing), which is: Brackets, Indicies, Division and Multiplication, Addition and Subtraction. So, brackets are computed first, then indices (powers), then division and multiplication have equal priority, and are done left...
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    What is the domain of f(x)=x^(2/3)?

    I don't think that a^{b/c} = \sqrt[c]{a^b} can be used when a is negative.
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    Pi and E combined See the fifth one down.
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    Proof of 1+1=2 ?

    There aren't any proofs if you keep on asking "why?", at some point or another you're going to have to accept things as true, without proof: these are axioms. Some common axioms for arithmetic are: 1) a + (b + c) = (a + b) + c 2) a + b = b + a 3) a + 0 = a 4) a + (-a) = 0 5) a*b = b*a 6)...
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    Proof of 1+1=2 ?

    It's proven, after hundreds of pages, in this book: I definitely can't explain how it's done, it's all in symbolic logic, and I guess you start with a degree level mathematical education.
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    Hard Geometry Challenge

    Could you do (y/1) = (1/x) instead?
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    Algebra inequalities and exponents

    For the first equation, your first step is wrong. Where you've attempted to subtract x from both sides, you've actually subtracted x from the left side, but only \frac{x}{4} from the left side. Adding, for example, 2 to a fraction, is not the same as adding 2 to the numerator of the fraction. In...
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    Algebra inequalities and exponents

    Your lack of brackets or LaTeX is both annoying and confusing. For your first question, the answer x < -3 only comes about if the equation is 2x + 5 < \frac{x-1}{4}, so I'm assuming that's what it is. From your working, it's a bit confusing as to what equation you're trying to solve...
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    Question regarding imaginary numbers and euler's formula

    Euler's formula says that e^{ix} = \cos x + i \sin x. Let x = 2 \pi n, where n is any integer. Then e^{ix} = \cos x + i \sin x = 1 + 0i = 1. Therefore, e^{ix} = 1 has an infinite number of solutions, all of the form x = 2 \pi n. The two you brought up, x=0 and x = 2 \pi are just the ones...
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    LaTeX Can I get some help on latex

    This is useful:
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    Mmmmmm Something very interesting I found about inequalities

    It's certainly easy, from a practical point of view, to put in the values first, so you know if you're multiplying/dividing by a negative number, which would change the inequality sign. Mathematically, if you consider separate cases for each variable being positive or negative, it makes no...
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    4 variable equation

    Multiply both sides by 1000.
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    Factorial identity question

    Nah, you've gone at that the wrong way. Start with the question, and use coolul007's hint to get a common denominator on the right hand side.