We are given two conditions: n is a natural number, and (n+1)/(2) is also a natural number. Natural numbers are typically positive integers (1, 2, 3,). Some definitions include 0 as a natural number, but for this proof, positive integers are sufficient.

We are given two conditions: n is a natural number, and (n+1)/(2) is also a natural number. Natural numbers are typically positive integers (1, 2, 3,). Some definitions include 0 as a natural number, but for this proof, positive integers are sufficient.

Since (n+1)/(2) is a natural number, we can represent it with a variable, say k. Setting k = (n+1)/(2) implies that k must be a natural number because the given statement tells us it is.

Since k is a natural number, its smallest possible value is 1 (assuming N = 1, 2, 3,). This means k is a positive integer.

To prove: $n \in \mathbb{N} \wedge \frac{n+1}{2} \in \mathbb{N} \Longrightarrow \exists m \in \mathbb{N} : n = 2m + 1$ . (natural numbers, implication, existence, odd number)

Answer: The statement $n \in \mathbb{N} \wedge \frac{n+1}{2} \in \mathbb{N} \Longrightarrow \exists m \in \mathbb{N} : n = 2m + 1$ is proven. By setting $k = \frac{n+1}{2}$ , we derived $n = 2k-1$ . Then, by defining $m = k-1$ , we showed that $n = 2m+1$ . Since $k \in \mathbb{N}$ , $k \ge 1$ , which implies $m = k-1 \ge 0$ . If $\mathbb{N}$ includes $0$ , then $m \in \mathbb{N}$ . If $\mathbb{N}$ starts from $1$ , then $m$ is a non-negative integer, which is consistent with the definition of an odd number.

Step-by-step solution

Step 1: Understand the Given Information

We are given two conditions: $n$ is a natural number, and $\frac{n+1}{2}$ is also a natural number. Natural numbers are typically positive integers $(1, 2, 3, \ldots)$ . Some definitions include $0$ as a natural number, but for this proof, positive integers are sufficient.

Step 2: Define a New Variable

Since $\frac{n+1}{2}$ is a natural number, we can represent it with a variable, say $k$ . Setting $k = \frac{n+1}{2}$ implies that $k$ must be a natural number because the given statement tells us it is.

Step 3: Rearrange the Equation

Now we rearrange the equation $k = \frac{n+1}{2}$ to express $n$ in terms of $k$ . First, multiply both sides by 2 to get $2k = n+1$ . Then, subtract 1 from both sides to isolate $n$ , resulting in $n = 2k - 1$ . This form is crucial for proving the final statement.

Step 4: Examine the Properties of k

Since $k$ is a natural number, its smallest possible value is 1 (assuming $\mathbb{N} = \{1, 2, 3, \ldots\}$ ). This means $k$ is a positive integer.

Step 5: Manipulate the Expression for n

We want to show that $n$ can be written in the form $2m+1$ for some natural number $m$ . We can rewrite $2k-1$ as $2(k-1)+1$ . This form directly relates to the target expression for $n$ .

Step 6: Define m and Verify it is a Natural Number

Let $m = k-1$ . Since $k$ is a natural number and $k \ge 1$ , then $m = k-1$ will be greater than or equal to $0$ . If $k=1$ , then $m=0$ . Some definitions of natural numbers include $0$ , while others do not. However, the standard definition of an odd number requires $m$ to be an integer (non-negative). If $n=1$ , then $1 = 2(0)+1$ , so $m=0$ . If the definition of natural numbers for $m$ includes $0$ , then $m \in \mathbb{N}$ . If it strictly means positive integers, this requires clarification. For example, in many contexts, $0$ is included in the set of non-negative integers or whole numbers, and $n=2m+1$ naturally leads to $m=0$ for $n=1$ . For the purpose of this proof, we will assume that the definition of natural numbers for $m$ allows for $0$ (i.e., $\mathbb{N} \cup \{0\}$ or $\mathbb{Z}_{\ge 0}$ ). If $m=0$ is not considered a natural number, then the statement would require $n>1$ for $m$ to be a 'natural' number.

Step 7: Conclude the Proof

By substituting $m = k-1$ into the expression for $n$ , we get $n = 2m+1$ . Since $k$ is a natural number, $m=k-1$ is a non-negative integer. If the definition of natural numbers includes zero, then $m \in \mathbb{N}$ . If not, then $m \in \mathbb{Z}_{\ge 0}$ . In either case, the form $n = 2m+1$ correctly defines $n$ as an odd number, which is what the implication means.

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