Question

4. Give arguments in support of the statement that there does not exist the largest natural number.
Hint.Every natural number has its successor​

Answer 1

Answer:

the following will prove what you are looking for:

Assume the existence of a largest natural number (p)

Being the largest natural number, 'p' cannot be followed by a greater natural number. Meaning:

p + 1 ≤ p

Now subtract 'p' from both sides. The result is:

1 ≤ 0

This has shown that the assumption of 'p' leads to a contradiction.

Therefore, it is not the case that a largest natural number exists.

QED

Here is a constructive proof.

Let P(n) be the property that there exists a natural number n′ where n′>n. We prove this by induction.

Base case - n=1: Clearly, P(1), since 484000>1.

Step – assume for n, prove for n+1: Assume by induction hypothesis that there exists an n′′ such that n′′>n. By the precongruence properties of > we easily see that n′′+484000>n+1. So take n′=n′′+484000.

If we want to prove the statement that it is not the case that there exists an n0∈N such that n≤n0 for all n∈N, then we assume that there exists such an n0. But then we have n0+484000≤n0, which is a contradiction. Therefore n0 cannot exist.

Answer 2

Step-by-step explanation:

sanjivansahara6

20.05.2019

Math

Secondary School

+5 pts

Answered

Give an argument in support of the statement that the largest natural number dose not exist

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YaBoi9897

YaBoi9897 Ambitious

the following will prove what you are looking for:

Assume the existence of a largest natural number (p)

Being the largest natural number, 'p' cannot be followed by a greater natural number. Meaning:

p + 1 ≤ p

Now subtract 'p' from both sides. The result is:

1 ≤ 0

This has shown that the assumption of 'p' leads to a contradiction.

Therefore, it is not the case that a largest natural number exists.

QED

Here is a constructive proof.

Let P(n) be the property that there exists a natural number n′ where n′>n. We prove this by induction.

Base case - n=1: Clearly, P(1), since 484000>1.

Step – assume for n, prove for n+1: Assume by induction hypothesis that there exists an n′′ such that n′′>n. By the precongruence properties of > we easily see that n′′+484000>n+1. So take n′=n′′+484000.

If we want to prove the statement that it is not the case that there exists an n0∈N such that n≤n0 for all n∈N, then we assume that there exists such an n0. But then we have n0+484000≤n0, which is a contradiction. Therefore n0 cannot exist.