Set of rational numbers $\mathbb{Q}$ is measure $0$. I approach this question by two sides. (First) Like here Showing that rationals have Lebesgue measure zero., $$ \mu(\mathbb Q) = \mu\left(\bigcup_{n=1}^\infty \{q_n\}\right) = \sum_{n=1}^\infty \mu(\{q_n\}) = \sum_{n=1}^\infty 0 = 0. $$ (Second) Using the definition of Lebuesgue measure. let’s order $\mathbb{Q}=\bigcup_{i=1}^{\infty}\left\{ r_{i}\right\} $. For given $\epsilon>0$, […]

I recently learned that $\cos{\theta} = \frac{e^{i\theta} + e^{-i\theta}}{2}$ and $\sin{\theta} = \frac{e^{i\theta} – e^{-i\theta}}{2}$ Based on this, I managed to “prove” that: $$e^{i\theta} = e^{-i\theta}$$ Since $e^{i\theta} = \cos{\theta} + i\sin{\theta}$, we can substitute the above two identities to get: $$e^{i\theta} = \frac{e^{i\theta} + e^{-i\theta}}{2} + i\frac{e^{i\theta} – e^{-i\theta}}{2}$$ Simplifying, I get $$(1-i)e^{i\theta} = […]

This is a widespread intuitive argument, asserting that $\mathrm{card} ( \mathbb{Q})=\mathrm{card}( \mathbb{Q^c})$: Between any two rational numbers there’s an irrational one and vice versa. So $\mathrm{card} ( \mathbb{Q})=\mathrm{card}( \mathbb{Q^c})$. How can one convince the learner that this argument is invalid?

I can hardly imagine an easier example of the fact that my understanding of the topic is more than rusty. I will divide the question in two parts to make the reading easier: 1) Background; 2) Problem. In the first part, I will show how I usually approach the proof that a sequence converge to […]

So I devised this proof that $1=0$. Of course it is false, but I don’t know why. Why? $$\begin{align*} x+1&=y\\ \frac{x+1}{y}&=1\\ \frac{x+1}{y}-1&=0\\ \frac{x+1}{y}-\frac{y}{y}&=0\\ \frac{x-y+1}{y}&=0\\ x-y+1&=0\\ x-y+1&=\frac{x-y+1}{y}\\ y(x-y+1)&=x-y+1\\ y&=1\\ x+1&=1\\ x&=0\qquad * * * *\\ y-1&=x\\ \frac{y-1}{x}&=1\\ \frac{y-1}{x}-1&=0\\ \frac{y-1}{x}-\frac{x}{x}&=0\\ \frac{y-x-1}{x}&=0\\ y-x-1&=0\\ y-x-1&=\frac{y-x-1}{x}\\ x(y-x-1)&=y-x-1\\ x&=1\qquad * * * *\\ 1&=0\\ \end{align*}$$

This question already has an answer here: Why $\sqrt{-1 \times {-1}} \neq \sqrt{-1}^2$? 8 answers

This question already has an answer here: $2+2 = 5$? error in proof 7 answers

I have searched for an answer to my question but no one seems to be talking about this particular matter.. I will use the all horses are the same color paradox as an example. Everyone points out that the statement is false for n=2 and that if we want to prove the propositions we should […]

I recently was working with square roots and came across this- $({\sqrt -1})$$=-1^\frac12$$=-1^\frac24$$=(-1^2)^\frac14$$=1^\frac14$$=1$ I understand that this is not true,but despite repeated attempts failed to prove it wrong.Can someone please point out the fallacy in this proof.Thanks in advance.

In David M. Burton’s book on Elementary Number Theory I have found the following words, … The first demonstrable progress toward comparing $\pi(x)$ with $\dfrac {x}{\ln x}$ was made by … P. L. Tchebycheff…, he proved that there exist positive constants $a$ and $b$ with $a$ $<$ $1$ $<$ $b$ such that- $$\dfrac{ax}{\ln x} < […]

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