2 votes

Do we not need to prove that Terence Tao's definition of the addition of natural numbers is true?

In principle you can define any notion as anything you want, as long as the definition meets formal criteria of making sense (like all symbols being used having been introduced prior, and them being ...
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2 votes

Proof addition is commutative in $ \mathbb R$

As both @Surb and @Violet Flame have mentioned in the comments, in order to formally prove properties of ${\mathbb{R}}$ you need first to give an actual formal construction of ${\mathbb{R}}$. Two ...
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1 vote

Proof addition is commutative in $ \mathbb R$

You can define real numbers as Cauchy sequences of rational numbers. Then the proposed property results from the commutativity from the rational numbers. You can check out the construction of the ...
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1 vote

Am I allowed to add and subtract expressions whose domain might not be $ \mathbb{R} $

Yes, but the validity of your solution is restricted to the range of $x$ that all your manipulations support. In your example, $x$ is restricted by the problem to $\Bbb R \setminus \{-\frac 12\}$ by ...
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1 vote

Solve for $x$ in $(x + (a<<s)-(b<<s))>>s = v$

$\def\l{\ll}$ $\def\r{\gg}$We have $a \l s = 2^sa$. The lower $s$ bits of $x$ don't matter. Thus we can represent $x$ as $$x = 2^s x_1 + x_0$$ with $0\leqslant x_0 < 2^s$. The equation is then $$\...
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1 vote

Which one is the larger : $20!$ or $2^{60}$?

$$20!=2^{18}\cdot3^8\cdot 5^4\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19$$ $$19\cdot 17\cdot 13=4199>2^{12}, \\3\cdot 11>2^5$$ So it suffices to show: $$3^7\cdot 5^4\cdot 7^2>2^{25}.$$ Now, $5\...
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1 vote

Which branch of math studies this problem?

Technically, you haven't reduced the number of computations, as you first needed to calculate $5 \times 10^6 + 7$ before squaring it, and must also find a way to extract the squares from the new ...
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