A course in mathematical analysis. - Derivatives and by Goursat E.

By Goursat E.

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1 (Multiplication of natural numbers). Let m be a natural number. To multiply zero to m, we define 0 × m := 0. Now suppose 30 2. Starting at the beginning: the natural numbers inductively that we have defined how to multiply n to m. Then we can multiply n++ to m by defining (n++) × m := (n × m) + m. Thus for instance 0 × m = 0, 1 × m = 0 + m, 2 × m = 0 + m + m, etc. By induction one can easily verify that the product of two natural numbers is a natural number. 2 (Multiplication is commutative). Let n, m be natural numbers.

3). 21. We should also caution that the subset relation ⊆ is not the same as the element relation ∈. The number 2 is an element of {1, 2, 3} but not a subset; thus 2 ∈ {1, 2, 3}, but 2 ⊆ {1, 2, 3}. Indeed, 2 is not even a set. Conversely, while {2} is a subset of {1, 2, 3}, it is not an element; thus {2} ⊆ {1, 2, 3} but {2} ∈ {1, 2, 3}. The point is that the number 2 and the set {2} are distinct objects. It is important to distinguish sets from their elements, as they can have different properties.

We shall use induction on n (keeping m fixed). , we show 0 + m = m + 0. 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).

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