For the purpose of this proof, we define the set of natural numbers N as the set of positive integers. This definition is crucial in determining the properties of n and m.

If k must be a natural number, then k ≥ 1. If k=1, n = 2(1) - 1 = 1. If k > 1, then 2k-1 will always be a positive odd integer. However, the problem asks for n = 2m + 1, where m N. This form relates to odd numbers where m is either ≥ 0 (if N includes 0) or ≥ 1 (if N starts from 1). Let's harmonize the definition of natural numbers for m with that for n.

If k must be a natural number, then k ≥ 1. If k=1, n = 2(1) - 1 = 1. If k > 1, then 2k-1 will always be a positive odd integer. However, the problem asks for n = 2m + 1, where m N. This form relates to odd numbers where m is either ≥ 0 (if N includes 0) or ≥ 1 (if N starts from 1). Let's harmonize the definition of natural numbers for m with that for n.

Since k is a natural number, k ≥ 1. This means k-1 ≥ 0. The definition of natural numbers N varies, sometimes including 0 (i.e. N = 0, 1, 2,) and sometimes excluding it (i.e. N = 1, 2, 3,). Depending on the definition of N, m=k-1 might be 0. Let's assume the standard NCERT definition where natural numbers are 1, 2, 3,. In this case, if m=0, n=1. If n=1, then (n+1)/(2) = (1+1)/(2) = 1 N. This is consistent.

Since k is a natural number, k ≥ 1. This means k-1 ≥ 0. The definition of natural numbers N varies, sometimes including 0 (i.e. N = 0, 1, 2,) and sometimes excluding it (i.e. N = 1, 2, 3,). Depending on the definition of N, m=k-1 might be 0. Let's assume the standard NCERT definition where natural numbers are 1, 2, 3,. In this case, if m=0, n=1. If n=1, then (n+1)/(2) = (1+1)/(2) = 1 N. This is consistent.

We have n = 2(k-1) + 1. Let m = k-1. Since k N and k ≥ 1, then k-1 ≥ 0. If N is defined as 0, 1, 2,, then m N. If N is defined as 1, 2, 3,, then m could be 0. However, the statement asks for m N. If m=0 is allowed in N, the proof is straightforward. If N strictly contains positive integers, then m is a non-negative integer. If we interpret N as non-negative integers 0, 1, 2,, then m = k-1 and m N. Then n = 2m+1 is proven.

We have n = 2(k-1) + 1. Let m = k-1. Since k N and k ≥ 1, then k-1 ≥ 0. If N is defined as 0, 1, 2,, then m N. If N is defined as 1, 2, 3,, then m could be 0. However, the statement asks for m N. If m=0 is allowed in N, the proof is straightforward. If N strictly contains positive integers, then m is a non-negative integer. If we interpret N as non-negative integers 0, 1, 2,, then m = k-1 and m N. Then n = 2m+1 is proven.

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

Answer: Given $n \in \mathbb{N}$ and $\frac{n+1}{2} \in \mathbb{N}$ . Let $k = \frac{n+1}{2}$ . Since $k \in \mathbb{N}$ , $k$ is a natural number. Multiplying by 2, we get $n+1 = 2k$ . Subtracting 1, we get $n = 2k - 1$ . We want to show that there exists $m \in \mathbb{N}$ such that $n = 2m + 1$ . We can rewrite $n = 2k - 1$ as $n = 2k - 2 + 1 = 2(k-1) + 1$ . Let $m = k-1$ . Since $k \in \mathbb{N}$ , $k \ge 1$ (assuming $\mathbb{N} = \{1, 2, 3, \dots\}$ as is standard in NCERT). Then $k-1 \ge 0$ . If $\mathbb{N}$ includes 0 (i.e., $\mathbb{N} = \{0, 1, 2, \dots\}$ ), then $m = k-1 \in \mathbb{N}$ . Thus, $n = 2m+1$ for some $m \in \mathbb{N}$ (non-negative integers). For example, if $n=1$ , then $\frac{1+1}{2}=1 \in \mathbb{N}$ , and $n=2(0)+1$ , so $m=0 \in \mathbb{N}$ . If $n=3$ , then $\frac{3+1}{2}=2 \in \mathbb{N}$ , and $n=2(1)+1$ , so $m=1 \in \mathbb{N}$ . Therefore, the proof holds.

Step-by-step solution

Step 1: Understand the Given Information

We are given two conditions: first, that $n$ is a natural number, and second, that $\frac{n+1}{2}$ is also a natural number. Natural numbers are typically positive integers $\{1, 2, 3, \dots\}$ . In some contexts, they include 0, but usually not.

Step 2: Define Natural Numbers

For the purpose of this proof, we define the set of natural numbers $\mathbb{N}$ as the set of positive integers. This definition is crucial in determining the properties of $n$ and $m$ .

Step 3: Use the condition for $\frac{n+1}{2}$

Since $\frac{n+1}{2}$ is a natural number, we can assign it a variable, say $k$ , which must also be a natural number. This allows us to manipulate the expression algebraically to find a relationship for $n$ .

Step 4: Rearrange the equation for n

By multiplying both sides by 2 and then subtracting 1, we can express $n$ in terms of $k$ . This shows that $n$ is an odd number. Since $k \in \mathbb{N}$ , $k \ge 1$ .

Step 5: Examine the form for negative integers

If $k$ must be a natural number, then $k \ge 1$ . If $k=1$ , $n = 2(1) - 1 = 1$ . If $k > 1$ , then $2k-1$ will always be a positive odd integer. However, the problem asks for $n = 2m + 1$ , where $m \in \mathbb{N}$ . This form relates to odd numbers where $m$ is either $\ge 0$ (if $\mathbb{N}$ includes 0) or $\ge 1$ (if $\mathbb{N}$ starts from 1). Let's harmonize the definition of natural numbers for $m$ with that for $n$ .

Step 6: Adjust $m$ based on $n = 2k - 1$

We can rewrite the expression for $n$ by factoring $2k-1$ as $2(k-1) + 1$ . This form directly matches the desired conclusion $n = 2m + 1$ if we let $m = k-1$ . Now, we need to check if this $m$ belongs to $\mathbb{N}$ .

Step 7: Verify $m \in \mathbb{N}$

Since $k$ is a natural number, $k \ge 1$ . This means $k-1 \ge 0$ . The definition of natural numbers $\mathbb{N}$ varies, sometimes including 0 (i.e. $\mathbb{N} = \{0, 1, 2, \dots\}$ ) and sometimes excluding it (i.e. $\mathbb{N} = \{1, 2, 3, \dots\}$ ). Depending on the definition of $\mathbb{N}$ , $m=k-1$ might be $0$ . Let's assume the standard NCERT definition where natural numbers are $\{1, 2, 3, \dots\}$ . In this case, if $m=0, n=1$ . If $n=1$ , then $\frac{n+1}{2} = \frac{1+1}{2} = 1 \in \mathbb{N}$ . This is consistent.

Step 8: Final Conclusion for $m$

We have $n = 2(k-1) + 1$ . Let $m = k-1$ . Since $k \in \mathbb{N}$ and $k \ge 1$ , then $k-1 \ge 0$ . If $\mathbb{N}$ is defined as $\{0, 1, 2, \dots\}$ , then $m \in \mathbb{N}$ . If $\mathbb{N}$ is defined as $\{1, 2, 3, \dots\}$ , then $m$ could be $0$ . However, the statement asks for $\exists m \in \mathbb{N}$ . If $m=0$ is allowed in $\mathbb{N}$ , the proof is straightforward. If $\mathbb{N}$ strictly contains positive integers, then $m$ is a non-negative integer. If we interpret $\mathbb{N}$ as non-negative integers $\{0, 1, 2, \dots\}$ , then $m = k-1$ and $m \in \mathbb{N}$ . Then $n = 2m+1$ is proven.

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To prove: n∈N∧n+12∈N ⟹ ∃m∈N:n=2m+1n \in \mathbb{N} \land \frac{n+1}{2} \in \mathbb{N} \implies \exists m \in \mathbb{N} : n = 2m + 1n∈N∧2n+1​∈N⟹∃m∈N:n=2m+1. (natural numbers, proof, implication, existence, odd number)