Question

Prove by contradiction that the set N of natural numbers is infinite

Answer 1

i gottta prove by contradiction

Answer 2

suppose N is finite. Since N is not equal to the null set, there exists a natural number k such that N is equivalent to N_k. Therefore, there exists a one-to-one function f from N_k onto N. Let n=max{f(1),f(2),...,f(n)}+1. Then n is not equal to f(i) for any element i of N_k. Therefore f is not onto N, a contradiction. Hence N is an infinite set.

Answer 3

To prove something by contradiction, you have to make an assumption and then show that that assumption leads to a contradiction. That would mean your initial assumption was wrong.

For this case, we want to prove that N is infinite. So, we'll make our assumption the opposite - that N is finite.

If N is finite, then we could take all the numbers in N and add them up, giving us X, a different number than any of the others in N. A property of the natural numbers is that when you add any of them together, you get another natural number. So what we have here is the number X wasn't in N, the list of all natural numbers, but X is a natural number because it's the sum of natural numbers. That's a contradiction.

Since we had a contradiction because we assumed that N is finite, then the opposite must be true. So, as a result, N is infinite.

Answer 4

Thhhankkksss both ur proofs were very clear

Answer 5

I have a question @succamc What does max{..} mean?

Answer 6

the largest value of the set

Answer 7

ohhhh gotcha. THANKKKSSSS GUYYSSSSS
ill medal u succamc and ull medal msalano