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By Terence Tao

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Extra resources for Analysis I (Volume 1)

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Because of the way we have defined 4 - it is the increment of the increment of the increment of the increment of 0 - it is not necessarily true a priori that this number is not the same as zero, even if it is "obvious". ("a priori" is Latin for "beforehand" - it refers to what one already knows or assumes to be true before one begins a proof or argument. 5, 4 was indeed equal to 0, and that in a standard two-byte computer representation of a natural number, for instance, 65536 is equal to 0 (using our definition of 65536 as equal to 0 incremented sixty-five thousand, five hundred and thirty-six times).

Proof. 2. 4 (Distributive law). For any natuml numbers a, b, c, we have a(b +c) = ab + ac and (b + c)a = ba +ca. Proof. Since multiplication is commutative we only need to show the first identity a(b + c) = ab + ac. We keep a and b fixed, and use induction on c. , a(b + 0) = ab + aO. The left-hand side is ab, while the right-hand side is ab + 0 = ab, so we are done with the base case. Now let us suppose inductively that a(b +c) = ab + ac, and let us prove that a(b + (c++)) = ab + a(c++ ). The left-hand side is a((b +c)++)= a(b+c) +a, while the right-hand side is ab+ac+a = a(b+c) +a by the induction hypothesis, and so we can close the induction.

5), for instance, is true. 5. (Indeed, one can give a "proof" of this fact. 5. Then P(O) is true, and if P(n) is true, then P( n++) is true. , no natural number can be a half-integer. 5 cannot be a natural number. 1. ) The principle of induction gives us a way to prove that a property P( n) is true for every natural number n. 11. A certain property P(n) is true for every natural number n. Proof. We use induction. , we prove P(O). (Insert proof of P(O) here). Now suppose inductively that n is a natural number, and P(n) has already been proven.

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