Understanding Mathematical Induction: Why Do We Use n Instead of k?

Hence, the use of n instead of k. This ensures that the proof holds true for any natural number, not just the specific value of k used in the original question.
  • #1
akshayacse
1
0
what is mathematical induction? explain with an example.
 
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  • #2
Mathematical Induction is a concept used to prove a given statement to be true for all positive integers. It is generally described with an example of effect of sequential falling of plates arranged parallel, up to a long distance. If one falls over the next, the whole sequence undergoes the same effect.
Theoretically it is done in three main steps:
1) Proving the statement to be true for the value 1
and hence showing the either sides of equation to be equal.
2) Proving for (k+1) value and implying on k-->(k+1)
3) Showing that the implication and the true value concludes the statement to be true for all n belonging to natural numbers.
An example:
Mathematical induction can be used to prove that the following statement, which we will call P(n), holds for all natural numbers n.
1+2+3+...+n = n(n+1)/2
P(n) gives a formula for the sum of the natural numbers less than or equal to number n. The proof that P(n) is true for each natural number n proceeds as follows.
Show that the statement holds for n = 1.
P(1) amounts to the statement:
1=1.(1+1)/2
In the left-hand side of the equation, the only term is 1, and so the left-hand side is simply equal to 1.
In the right-hand side of the equation, 1·(1 + 1)/2 = 1.
The two sides are equal, so the statement is true for n = 1. Thus it has been shown that P(1) holds.
Show that if P(k) holds, then also P(k + 1) holds. This can be done as follows.

Assume P(k) holds (for some unspecified value of k). It must then be shown that P(k + 1) holds, that is:
1+2+...+k+(k+1) = (k+1)[(k+1)+1]/2

Using the induction hypothesis that P(k) holds, the left-hand side can be rewritten to:
[k(k+1)/]+(k+1)

Algebraically it is equal to (k+1)[(k+1)+1]/2
thereby showing that indeed P(k + 1) holds.

Since both the basis and the inductive step have been proved, it has now been proved by mathematical induction that P(n) holds for all natural n.
 
Last edited:
  • #3
PhysicoRaj said:
Mathematical Induction is a concept used to prove a given statement to be true for all positive integers. It is generally described with an example of effect of sequential falling of plates arranged parallel, up to a long distance. If one falls over the next, the whole sequence undergoes the same effect.
Theoretically it is done in three main steps:
1) Proving the statement to be true for the value 1
and hence showing the either sides of equation to be equal.
2) Proving for (k+1) value and implying on k-->(k+1)
3) Showing that the implication and the true value concludes the statement to be true for all n belonging to natural numbers.
An example:
Mathematical induction can be used to prove that the following statement, which we will call P(n), holds for all natural numbers n.
1+2+3+...+n = n(n+1)/2
P(n) gives a formula for the sum of the natural numbers less than or equal to number n. The proof that P(n) is true for each natural number n proceeds as follows.
Show that the statement holds for n = 1.
P(1) amounts to the statement:
1=1.(1+1)/2
In the left-hand side of the equation, the only term is 1, and so the left-hand side is simply equal to 1.
In the right-hand side of the equation, 1·(1 + 1)/2 = 1.
The two sides are equal, so the statement is true for n = 1. Thus it has been shown that P(1) holds.
Show that if P(k) holds, then also P(k + 1) holds. This can be done as follows.

Assume P(k) holds (for some unspecified value of k). It must then be shown that P(k + 1) holds, that is:
1+2+...+k+(k+1) = (k+1)[(k+1)+1]/2

Using the induction hypothesis that P(k) holds, the left-hand side can be rewritten to:
[k(k+1)/]+(k+1)

Algebraically it is equal to (k+1)[(k+1)+1]/2
thereby showing that indeed P(k + 1) holds.

Since both the basis and the inductive step have been proved, it has now been proved by mathematical induction that P(n) holds for all natural n.

Why do we have to turn k into n? Why can't we just use k the whole time since that is what we use in the original question?
 
  • #4
Well, it is a purely technical issue: the property holds for the natural number 1

(tho this is not always necessary), and we denote that it is satisfied by the natural

number n=k, and we want this last to imply that the property holds also for the

proposition P(k+1), which is indexed by the natural number n=k+1.

Look up too, the well-ordering principle , which is equivalent to induction

--we use the fact that in standard induction, the propositions are indexed by the

natural numbers with their standard ordering. You may also want to look up

transfinite induction , which generalizes standard induction to well-ordered sets
 
  • #5
student34 said:
Why do we have to turn k into n? Why can't we just use k the whole time since that is what we use in the original question?
P(n) is the general form. The proof is given to show both for values of 1 (sometimes it is also proved for 0) and any other natural number k and k+1, the statement holds true, which for the sake of convenience you can do with n+1 and n.
 

1. What is mathematical induction?

Mathematical induction is a proof technique used in mathematics to prove that a statement is true for all natural numbers. It involves breaking down a statement into smaller cases and proving that it holds for the smallest case (usually 0 or 1) and then showing that if it holds for a certain case, it must also hold for the next case.

2. How does mathematical induction work?

Mathematical induction works by showing that a statement is true for the smallest case, usually 0 or 1. Then, it is assumed that the statement holds for some arbitrary case, and using this assumption, it is proven that the statement also holds for the next case. This process is repeated until it can be shown that the statement holds for all natural numbers.

3. What are the steps involved in a mathematical induction proof?

The steps involved in a mathematical induction proof are as follows:

  1. Base case: Prove that the statement holds for the smallest case, usually 0 or 1.
  2. Inductive hypothesis: Assume that the statement holds for some arbitrary case.
  3. Inductive step: Use the inductive hypothesis to prove that the statement also holds for the next case.
  4. Conclusion: Conclude that the statement holds for all natural numbers.

4. What types of statements can be proven using mathematical induction?

Mathematical induction is typically used to prove statements that involve natural numbers, such as equations, inequalities, and divisibility. It can also be used to prove statements about sequences and series.

5. Are there any limitations to using mathematical induction?

Yes, there are some limitations to using mathematical induction. It can only be used to prove statements that involve natural numbers, and it may not be the most efficient or straightforward method of proof for certain statements. Additionally, it is important to ensure that the base case and inductive step are both valid and logically sound in order for the proof to be considered valid.

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