To prove: n∈N∧2n+1∈N⟹∃m∈N:n=2m+1. (natural numbers, proof, implication, existence, odd number)
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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 2n+1 is also a natural number. Natural numbers are typically positive integers {1,2,3,…}. 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 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 2n+1
Since 2n+1 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∈N, k≥1.
Step 5: Examine the form for negative integers
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.
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 N.
Step 7: Verify m∈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 2n+1=21+1=1∈N. This is consistent.
Step 8: Final Conclusion for m
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.