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Hello people.

I'm a freshman in college and I'm looking forward to enter an advanced mathematics program. The admission test is basically number theory & sometimes I get stuck with some questions (I'm using Spivak's Calculus to study).

I'm stuck on question 8, chapter 2.

"Prove that every natural number is either odd or even."

I've come to a solution though I don't think it is right and it doesn't seem very elegant. I'm doing this by induction.

Let X be a non-empty set of natural numbers that are not even or odd.

Let A an specific set of natural numbers that are not odd, so they must be even.

An specific number [tex]n[/tex] is even if, and only if, it satisfies the equation [tex]n=2k,~ \forall k \in \mathbb{Z}[/tex] and odd if, and only if, it satisfies the equation [tex]n=2k +1 ,~ \forall k \in \mathbb{Z}[/tex]

So if A is a non-empty set, it must have a smallest number [tex]y=2x,~ \forall x \in \mathbb{Z}[/tex]

[tex]A(x) = 2x [/tex]

[tex]A(x) + 1 = 2x + 1[/tex]

This way, we proved that the first natural number following an even is odd.

Let B be an specific set of natural numbers that are not even, so it must be odd.

So if B is a non-empty set, it must have a smallest number [tex]i=2j +1,~ \forall j \in \mathbb{Z}[/tex]

[tex]B(j) = 2j +1[/tex]

[tex]B(j) + 1 = 2j + 2 = 2(j+1)[/tex]

Therefore, the smallest natural number after an odd is even. So [tex]A \cup B = \mathbb{N}[/tex] therefore [tex]X = \varnothing[/tex], which contradicts a previous statement, therefore every natural number is either odd or even.

I'm a freshman in college and I'm looking forward to enter an advanced mathematics program. The admission test is basically number theory & sometimes I get stuck with some questions (I'm using Spivak's Calculus to study).

I'm stuck on question 8, chapter 2.

"Prove that every natural number is either odd or even."

I've come to a solution though I don't think it is right and it doesn't seem very elegant. I'm doing this by induction.

Let X be a non-empty set of natural numbers that are not even or odd.

Let A an specific set of natural numbers that are not odd, so they must be even.

An specific number [tex]n[/tex] is even if, and only if, it satisfies the equation [tex]n=2k,~ \forall k \in \mathbb{Z}[/tex] and odd if, and only if, it satisfies the equation [tex]n=2k +1 ,~ \forall k \in \mathbb{Z}[/tex]

So if A is a non-empty set, it must have a smallest number [tex]y=2x,~ \forall x \in \mathbb{Z}[/tex]

[tex]A(x) = 2x [/tex]

[tex]A(x) + 1 = 2x + 1[/tex]

This way, we proved that the first natural number following an even is odd.

Let B be an specific set of natural numbers that are not even, so it must be odd.

So if B is a non-empty set, it must have a smallest number [tex]i=2j +1,~ \forall j \in \mathbb{Z}[/tex]

[tex]B(j) = 2j +1[/tex]

[tex]B(j) + 1 = 2j + 2 = 2(j+1)[/tex]

Therefore, the smallest natural number after an odd is even. So [tex]A \cup B = \mathbb{N}[/tex] therefore [tex]X = \varnothing[/tex], which contradicts a previous statement, therefore every natural number is either odd or even.

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