Transcendental Numbers - Numberphile
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users cite it as an accessible description of what makes a number algebraic versus transcendental, both when asking about the idea and when explaining it to others.
Recommended by 1 people across 1 different months, 2013 to 2016.
What it is
A Numberphile episode in which Australia's Numeracy Ambassador Simon Pampena explains transcendental numbers, contrasting them with algebraic numbers and using e and pi as examples. It is an informal, friendly explanation filmed by Brady Haran.
What people said
Recommends Eminem · 2015-06 · Mathematics Stack Exchange · on “Wrong proof...But where is the mistake?”
So I've just watched this wonderful Numberphile [this video] video about transcendental numbers. In the video, the guy shows that $$e=\sum_{n=0}^\infty\frac{1}{n!}=1+\frac{1}{1}+\frac{1}{1\cdot2}+\frac{1}{1\cdot2\cdot3}+\cdots$$ In the video, he says that if a number can be reduced to zero, it is algebraic. Now, if we take: $$1+\frac{1}{1}+\frac{1}{1\cdot2}+\frac{1}{1\cdot2\cdot3}+\cdots$$ and multiply that by 1, we can get: $$1+1+\frac{1}{2}+\frac{1}{2\cdot3}+\cdots$$ take that and multiply by $2$ , we get: $$2+2+1+\frac{1}{3}+\cdots$$ Next we will get: $$6+6+3+1+\cdots$$ And so on. Wait, isn't that something that can be reduced to zero? Obviously there is a mistake somewhere, but I can't seem t
Cites Ben Grossmann · 2013-08 · Mathematics Stack Exchange · on “Is there a proof that says that an operation that can take a transcendental number and make it an integer cannot exist?”
You're a little bit off with your definition of transcendental numbers. The "algebraic numbers" (i.e. the "non-transcendental" numbers) can be thought of as follows. We begin by considering the numbers that can be brought to zero under the following operations: addition by an integer multiplication by a (nonzero) integer (positive) integer powers Any such number and any sum of finitely many of these numbers is algebraic. If a number isn't algebraic, it's transcendental. $3$ is algebraic because $3-3=0$ . $\frac34$ is algebraic because $4\times \frac34 = 3$ , and $3-3=0$ . $\sqrt3$ is algebraic because $(\sqrt3)^2=3$ , and $3-3=0$ . $e$ is transcendental because no amount or order of these op
Cites alan2here · 2016-07 · Mathematics Stack Exchange · on “Are irrationals with an increasingly padded infinite series in the decimals non-algebraic?”
I should have mentioned that the operations must be hyper-operations or there inverses {+, -, *, /, etc...}, see this numberphile episode [this video] I like your description but it's not that apparent to me why thats the case. They phrase it differently but it means the same thing, which is that using a finite number of these aloud operations and whole numbers, can you bring the number to 0. For example with x = 1/sqrt(5) then (1/x)^2 - 5 = 0 so 'x' is algebraic.