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The spam links in Mathegocart’s quote should be removed as well.
Last night I dreamed that my house was suddenly invaded by cats – which all came in through a gap in the wall of the bathroom.
I think Cantor made a real mess of this field. In mapping the counting numbers (proportional to some finite limit N) to the rational numbers (proportional to the finite limite N²) he made infinity impossible to understand in logical terms. Because it is infinity, that doesn't make everything magically correct. You should be able to think of infinity as some value N which is CONSISTENTLY increased without restriction. He insisted on increasing N inconsistently and therefore came up with some rather foolish results.
Here is how I do it:
http://lesliegreen.byethost3.com/articles/new_maths.pdf
I sympathize with you; I myself have reservations about Cantor’s diagonal proof of the uncountability of the real numbers. For example, he seemed to assume that there was a bijection between each real number as its infinite decimal representation – this is not quite true, as seen by the fact that
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Here is how Jade Tan-Holmes does it on her YouTube channel Up and Atom: An Alternative Proof That The Real Numbers Are Uncountable.
I recently started learning about partial differential equations. The video gives the heat equation as an example of a PDE – and includes a fascinatingly watchable derivation of the equation discovered by Joseph Fourier.
Another example of a PDE is the wave equation:
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hi;
If there is an equation satisying the relation:
and
How to prove that
is composite?
That is not true. Take, for example, n = 3, x = 5, m = 5, c = 3. Here, nx = mc and gcd(m,n) = 1 but c=3 is prime.
6. The quadratic equation whose roots are squares of the roots of is x.
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This game: https://www.mathsisfun.com/games/master-detective.html.
Is there a glitch in the software?

If there is a glitch, please fix it. I’ve just wasted four hours of my morning trying to solve an unsolvable game.
Thanks.
JaneFairfax wrote:UOIAE
fluoridate
fluorinate
subordinate
uncomplicated
unprofitable
unsociableunmotivated
glucosidase
Q: What do you take when you have a headache?
A: Take nothing – because nothing acts faster than Anadin!
Similarly, if you want to be perfect, be nobody – because nobody is perfect!
Sir Michael Francis Atiyah … is a British-Lebanese mathematician specialising in geometry.
At a hotly-anticipated talk at the Heidelberg Laureate Forum today [Monday 24 September 2018], retired mathematician Michael Atiyah delivered what he claimed was a proof of the Riemann hypothesis, a challenge that has eluded his peers for nearly 160 years.
If a = 2, b = 1, c = 4, then A = √(2+1·√ 4) = 2, B = √(2−1·√ 4) = 0, S = A + B = 2 ≠ b.
Draw three distinct, parallel lines and another line (a transversal) to cross them all.
But those three lines may not be in one and the same plane.
Let x be the number of people enrolled in all three classes. Then:
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[*]170−x are enrolled in art and drama but not piano,[/*]
[*]150−x are enrolled in drama and piano but not art,[/*]
[*]300−x are enrolled in piano and art but not drama.[/*]
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Now that you have split the numbers into non-overlapping groups, you can add them up:
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Let us apply zetafunc’s method in another way. Dividing by 2 and rearranging gives
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Now consider the following table:
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Thus we see that outside of {0,±1} the equation has no solution as the LHS and RHS have opposite signs. This proves that there are no solutions other than x = 0, ±1.
There are precisely three solutions:
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Unfortunately I do not know how to prove it rigorously. Maybe Bob Bundy can come up with more helpful information.
Here’s a YouTube video on tests of divisibility and how to apply them to some math problems.
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I found the solution to finding all palindromic four-digit powers particularly interesting.
Can you help me solve this equation?
What do you mean by why is n term not 2n−1?
Since the thread title is mathematical induction, I take it you want to prove the formula for the sum of the first n odd positive integers. To do this, you need to know what the formula is in the first place. If you’re not given the formula, you can make a guess and then proceed to prove it. So, trying the first few sums …
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I think you can make a guess as to what the formula is going to be.
I presume ɑ and β are the roots, which are complex. Since we are going to deal with their powers, best to convert them to cos–sin form:
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So the roots are
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one of them is ɑ, the other is β. Since the question doesn’t say which is which, you may as well take your pick. Now use De Moivre’s theorem.
Why can’t they use metric?
Do the long division yourself!