Proving Inductive Step: P(k) to P(k+1)

  • Thread starter Chromium
  • Start date
In summary, this conversation is discussing the proof of the statement P(n): 1 + 3 + 5 + ... + (2n-1) = n^2 and how to prove it using induction. The conversation includes the steps of proving P(1), the induction hypothesis P(k), and the inductive step P(k+1). It is mentioned that the inductive step may differ from problem to problem, but the key is to use the induction hypothesis to prove P(k+1).
  • #1
Chromium
56
0
P(n): 1 + 3 + 5 + ... + (2n-1) = n^2

Prove P(1)
P(1) = 1 = 1^2
P(1) is true

(A) P(k): 1 + 3 + 5 +...+ (2k-1) = k^2

(B) P(k + 1): 1 + 3 + 5 +...+ (2k-1) + (2(k+1)-1)
or
(2k-1) + (2k + 1) = (k+1)^2

Assuming A, prove B

(2k-1) = k^2
(2(k+1)-1) = (k+1)(k+1)
(2k + 1) = (k^2 + 2k+ 2)
= (k+1)^2

When in comes to the inductive step, does it "differ" from problem to problem? I can always get to the inductive assumption, but then I'm never sure just how to go about proving it.
 
Physics news on Phys.org
  • #2
Chromium said:
P(n): 1 + 3 + 5 + ... + (2n-1) = n^2

Prove P(1)
P(1) = 1 = 1^2
P(1) is true

(A) P(k): 1 + 3 + 5 +...+ (2k-1) = k^2

(B) P(k + 1): 1 + 3 + 5 +...+ (2k-1) + (2(k+1)-1)
or
(2k-1) + (2k + 1) = (k+1)^2
I'm, sorry? how does 1+ 3+ ...+ 2(k+1)= (2k-1)+ 2k?

Assuming A, prove B

(2k-1) = k^2
(2(k+1)-1) = (k+1)(k+1)
(2k + 1) = (k^2 + 2k+ 2)
= (k+1)^2

When in comes to the inductive step, does it "differ" from problem to problem? I can always get to the inductive assumption, but then I'm never sure just how to go about proving it.
You want to prove that "P(k)= (k+1)^2. P(k+1)= P(k)+ 2k+1 for every k. The "induction hypothesis" is that P(k-1)= k^2 . From that, P(k+1)= P(k)+ k+1= k^2+ 2k+ 1= (k+1)^2/
 
  • #3


I understand that the inductive step may differ from problem to problem. It is important to carefully examine the given statement and determine what needs to be proven. In this case, the inductive step is proving that P(k) implies P(k+1). This means that by assuming P(k) is true, we must show that it leads to P(k+1) being true as well. In this specific problem, we use algebraic manipulation to show that P(k+1) is equivalent to P(k) being true. In other problems, different methods may need to be used. It is important to carefully analyze the given statement and use logical reasoning to determine the best approach for proving the inductive step.
 

1. What is the purpose of proving the inductive step in a mathematical proof?

The inductive step is used to show that a statement, P(k), is true for all values of k. This is necessary in mathematical proofs to establish the validity of a statement or theorem for all cases.

2. How is the inductive step different from the base case?

The base case is used to prove that the statement, P(1), is true for the first value of k. The inductive step then shows that if P(k) is true, then P(k+1) must also be true. This combination of the base case and inductive step proves the statement for all values of k.

3. Can the inductive step be skipped in a proof?

No, the inductive step is a necessary part of proving the validity of a statement or theorem for all values of k. Without it, the proof would not be complete and the statement would not be proven to be true for all cases.

4. How do you know if the inductive step is valid?

The inductive step is considered valid if it follows the proper logical reasoning and uses the correct assumptions and mathematical operations. It is important to carefully follow the steps and make sure each step is logically connected to the previous one.

5. Are there any common mistakes when proving the inductive step?

Yes, some common mistakes include not providing enough detail in the inductive step, making incorrect assumptions, or using faulty logic. It is important to carefully check each step and make sure it is logically sound before moving on to the next step.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
  • Precalculus Mathematics Homework Help
Replies
1
Views
728
  • Precalculus Mathematics Homework Help
Replies
17
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
  • Precalculus Mathematics Homework Help
Replies
15
Views
1K
  • Precalculus Mathematics Homework Help
Replies
8
Views
1K
  • Precalculus Mathematics Homework Help
Replies
13
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
4K
  • Precalculus Mathematics Homework Help
Replies
10
Views
736
Back
Top