# Tag Info

3

Standard construction is: $0:=\varnothing$ $n+1:=n\cup\{n\}$ Order $<$ is actually the same as order $\in$. Then $1=\{0\}$ and $2=\{0,1\}$ so that $1\in2$ or equivalently $1<2$.

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The usual answer is to use Newton's method, which preserves rational numbers: $$x=\frac12\left(w+\frac2w\right)$$ Rudin's book contains a slightly different expression. See here.

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Hint: If $p-q\sqrt{2}>0$ for $p,q$ positive integers, multiply by $0<3-2\sqrt{2}<1$.

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This is true for any rational number $p$. If rational $x$ satisfies $x^2 < p$, then, since the rationals are dense, there is a rational $q$ such that $x < q < \sqrt{p}$, whether or not $p$ is a perfect squate. Then $x^2 < q^2 < p$.

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The way I like to approach these is to start by asking: what numbers are we given? We're given $0$ through $4$ - what does that tell us about our base? We should first consider $4\times 3$ as we do multiplication from right to left. Since $4\times 3 = 12$ and $12 \equiv 2 \pmod b$ by equating the ones digits in your equation, you know that \$12-2\equiv 0 ...

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