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(Pn): For all r with 1 ≤ r < n, there is no one-to-one function from Nn to Nr.

Let Nk= {1,2,…,k} be a finite set with k elements. This problems shows that if n, r elements of N with n > r, then there is no one-to-one function from Nn to Nr. We prove this by induction on n = 2 for the following statement.

(Pn): For all r with 1 = r < n, there is no one-to-one function from Nn to Nr.

a. The first step covers the general case when r=1 and any n>r.

Prove that if n > r=1, then there is no one-to-one function from Nn to Nr.

We will take: Quote:

c. Assume the induction hypothesis (Pn) is true and prove that (Pn+1) is true.

Hint: If r = 2 and f: Nn+1 to Nr were one-to-one, then the restriction of f to Nn into Nr \ {f(n+1)} would be one-to-one. You may also use the fact that there is a bijection from Nr \{f(n+1)} to Nr-1.

Suppose that for some $\displaystyle n \geq 2$ P(n) is true and that P(n+1) is false. Quote:

e. Prove that if n,r is an element of N with n not = r, then Nn does not have the same cardinality as Nr.

Two sets $\displaystyle A$ and $\displaystyle B$ have the same cardinality if there exits a one-to-one