Proving InfE <= SupE, Proving 1/n < a < n, and Proving N = E

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1. Let E be a nonempty subset of R (real numbers)

Prove that infE <= supE

2. Prove that if a > 0 then there exists n element N (natural) such that 1/n < a < n

3. A subset E of te real numbers R is an inductive set if

i) 1 element E
ii) If x element E then x + 1 element E

A real number is called a natural number if it belongs to every inductive set. The set of natural numbers is denoted by N. Recall that the principle of mathematical inductions says that if M is any subset of N that is an inductive set then M = N. Show that N = E, where E = {1,2,3,4...}

Any help would be greatly appreciated ! :)
 
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Please show us your attempt at a solution.
 
i can not some up with anything i am not sure where to start for any of them
 
Do you know the definitions of inf A and sup A ?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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