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### A challenge by R. P. Feynman: give counter-intuitive theorems that can be translated into everyday language

The following is a quote from Surely you're joking, Mr. Feynman . The question is: are there any interesting theorems that you think would be a good example to tell Richard Feynman, as an answer to ...
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### Which one result in mathematics has surprised you the most? [closed]

A large part of my fascination in mathematics is because of some very surprising results that I have seen there. I remember one I found very hard to swallow when I first encountered it, was what is ...
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### The Monty Hall problem

I was watching the movie 21 yesterday, and in the first 15 minutes or so the main character is in a classroom, being asked a "trick" question (in the sense that the teacher believes that he'll get the ...
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### Proving that 1- and 2-d simple symmetric random walks return to the origin with probability 1

How does one prove that a simple (steps of length $1$ in directions parallel to the axes) symmetric (each possible direction is equally likely) random walk in $1$ or $2$ dimensions returns to the ...
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### Examples of results failing in higher dimensions

A number of economists do not appreciate rigor in their usage of mathematics and I find it very discouraging. One of the examples of rigor-lacking approach are proofs done via graphs or pictures ...
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Consider the curve $\frac{1}{x}$ where $x \geq 1$. Rotate this curve around the x-axis. One Dimension - Clearly this structure is infinitely long. Two Dimensions - Surface Area = $2\pi\int_∞^1\... 2answers 368 views ### Constructing a number not in$\bigcup\limits_{k=1}^{\infty} (q_k-\frac{\epsilon}{2^k},q_k+\frac{\epsilon}{2^k})$I have couple of questions related to the properties of real numbers. My first question is as follows. Let$S_{\epsilon} = \displaystyle \bigcup_{k=1}^{\infty} \left( q_k-\frac{\epsilon}{2^{k+1}},...
From Baire category theorem, we see that $\mathbb{Q}$ can not be a $G_{\delta}$. But consider the following construction: Let us consider $\mathbb{Q}\cap [0,1]$, putting all the elements in the set ...