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By Terence Tao

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We shall use induction on n (keeping m fixed). , we show O+m = m+O. 2, m + 0 = m. Thus the base case is done. Now suppose inductively that n+m = m+n, now we have to prove that (n++) + m = m + (n++) to close the induction. By the definition of addition, (n++) +m = (n+m)++. 3, m + (n++) = (m + n)++, but this is equal to (n + m)++ by the inductive hypothesis n + m = m + n. Thus (n++) + m = m + (n++) and we have closed the induction. 5 (Addition is associative). For any natural numbers a, b, c, we have (a+ b)+ c =a+ (b +c).

G. ). 1. 4. , ifn, m are natural numbers and n f=. m, then n++ f=. m+t-. l, if n++ = m-t~t, then we must haven= m. 8. 6 is not equal to 2. Proof. Suppose for sake of contradiction that 6 = 2. 4 we have 5 = 1, so that 4++ = 0++. 4 again we then have 4 = 0, which contradicts our previous proposition. D As one can see from this proposition, it now looks like we can keep all of the natural numbers distinct from each other. 2) allow us to confirm that 0, 1, 2, 3, ... 9. 5, ... }. 4 are still satisfied for this set.

The analysis you learn in this text will help you resolve these questions, and will let you know when these rules (and others) are justified, and when they are illegal, thus separating the useful applications of these rules from the nonsense. Thus they can prevent you from making mistakes, and can help you place these rules in a wider context. Moreover, as you learn analysis you will develop an "analytical way of thinking", which will help you whenever you come into contact with any new rules of mathematics, or when dealing with situations which are not quite covered by the standard rules, For instance, what if your functions are complex-valued instead of real-valued?

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