Question

Write a proposition and then write a proof of the proposition. The proposition should be in the form: If m is a natural number and m ≥ 2, then...

a) If m is a natural number, then m + 2 is also a natural number
b) If m is a prime number, then m² + 1 is also prime
c) If m is an even number, then m/2 is an integer
d) If m is an odd number, then 2m is an even number

Answer 1

The proposition that if m is an odd number, then 2m is an even number is proven using the definition of odd numbers and mathematical manipulation.

Step-by-step explanation:

The proposition we'll explore is: If m is a natural number and m ≥ 2, then if m is an odd number, then 2m is an even number. We will write a proof of this proposition following mathematical logic.

Proof of the Proposition

First, let's define an odd number. An odd number m is defined as any natural number that can be expressed in the form m = 2k + 1, where k is an integer. Now, if m is an odd number, we can express it as 2k + 1 for some integer k.

To prove that 2m is an even number, we proceed as follows:

Let m be an odd number, so m = 2k + 1 where k is an integer.

Multiply m by 2 to get 2m, so 2m = 2(2k + 1).

Simplify the equation: 2m = 4k + 2 = 2(2k + 1).

Notice that 2m can be written as 2 times another integer (2k + 1), which defines it as an even number.

Therefore, we have shown that if m is an odd number, then 2m must be an even number.