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I assume the equivalence relation $\sim$ is "there exists a bijective map between". Then neither does $A\sim X$ have anything to do with $a\sim x$ for $a\in A$, $x\in X$ (for example $\{\mathbb N\}$ and $\{\mathbb R\}$ are both one-element sests and yet theri only elements have different cardinalitty). Nor does the fact that some injective map $A\to X$ ...


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The logical error is the conclusion of the two proofs. Your specific definitons of a chain and an antichain together do not cover all the possibilities, so "is not a chain" does not necessarily imply "is an antichain." Using the specific definitions you've given would imply that $\emptyset$ is neither a chain nor an antichain. If you changed the definition ...


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Suppose there are $n$ elements $s_i \in S$ and the largest is $M$. Then looking at the mean we have $\dfrac{\sum_i s_i} {n}=N$, and each $s_i \le M$, so $ nN=\sum_i s_i \le \sum_i M=nM$ meaning $nN \le nM$ and so $$N \le M$$ i.e. the mean is less than or equal to the maximum. No need to look for a contradiction.


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Enumerate the elements of $S$ as $S=\{s_1,\ldots,s_K\}$. Suppose that it was true that $s_i<N$ for all $i$. Then, $$ \sum_{i=1}^K s_i<\sum_{i=1}^K N=K\times N=K\frac{\sum_{i=1}^K s_i}{K}=\sum_{i=1}^K s_i. $$ The leftmost term and rightmost term are the same so you have a contradiction.


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Answer to Question 1: Yes, that is one version of the most commonly given ("Pythagorean") proof of the irrationality of $\sqrt 2$. Answer to Question 2: Using the proof you give, you are constructing two infinite, descending sequences $p_i,q_i$ of natural numbers by repeatedly extracting their common factor $2$. Given that there are only finitely many ...


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What is a rational number? The definition I use is that it's a number that can be written as a ratio of two integers. If your book doesn't say explicitly that the integers can be coprime, it's because the notion of a reduced fraction is fundamental to working with fractions. You're not allowed to work with fractions until you believe that they can be written ...



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