Questions tagged [induction]

For questions about mathematical induction, a method of mathematical proof. Mathematical induction generally proceeds by proving a statement for some integer, called the base case, and then proving that if it holds for one integer then it holds for the next integer. This tag is primarily meant for questions about induction over natural numbers but is also appropriate for other kinds of induction such as transfinite, structural, double, backwards, etc.

Mathematical induction is a form of deductive reasoning. Its most common use is induction over well-ordered sets, such as natural numbers or ordinals. While induction can be expanded to class relations which are well-founded, this tag is aimed mostly at questions about induction over natural numbers.

In general use, induction means inference from the particular to the general. This is used in terms such as inductive reasoning, which involves making an inference about the unknown based on some known sample. Mathematical induction is not true induction in this sense, but is rather a form of proof.

Induction over the natural numbers generally proceeds with a base case and an inductive step:

  • First prove the statement for the base case, which is usually $n=0$ or $n=1$.
  • Next, assume that the statement is true for an input $n$, and prove that it is true for the input $n+1$.

The following variant goes without a base case: Assuming the statement is true for all $n\in\mathbb N$ with $n < N$, prove that is true for $N$, too. This has to be done for all $N\in\mathbb N$.

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Prove using induction that, $|\{B ⊆ A_n \mid 0 ∈ B\}| = 2^{n−1}$ where $A_n = \{k ∈ N \mid k < n\}$ and $n ≥ 1$

So far I've found that the $n = 1$ is the base case. However, I can't seem to progress to actually proving his. If the set $B ⊆ A_n$ do I have to prove that $2^{n-1}$ applies to both the empty and singleton set (all possible subsets)? I'm also…
Habib C
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can the method of principle of mathematical induction be applied to prove or disprove all statements involving natural numbers?

I was solving the following question Let $S(k) = 1+3+5+\cdots+[2k-1] = 3+k^2$. Then which of the following is true ? $S(1)$ is correct $S(k) \implies S(k+1)$ $\neg (S(k) \implies S(k+1))$ Principle of mathematical induction can be used to prove…
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Proof by Induction $2\cdot 7^n + 3\cdot 5^n - 5$ is multiple of $24$

Proof by induction that $2\cdot 7^n + 3\cdot 5^n - 5$ is a multiple of $24$. I tried solving but got stuck. Show that it is true for n=1 $$2\cdot 7^1 + 3\cdot 5^1 - 5 = 14 + 15 - 5 = 24$$ Assume it true for $n = k$ $$2\cdot 7^k + 3\cdot 5^k - 5 =…
Drex
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Show that $2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! = n(n + 1)!$

Show that $$2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! = n(n + 1)!$$ I got stuck after the n+1 part. $$((n+1)^2 + 1)(n+1)! = (n+1)(n+2)!.$$ I'm not sure how to proceed from this part.
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Inductive Step For a Summation mathematical induction

I have a bit of trouble with how I should go about showing the inductive step for my induction problem. I know the general idea is to show that it can work for all numbers based on the base case but I'm still stuck on how to show it. $\sum_{i=2}^n…
Elchavo18
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Why is this Induction Proof Wrong?

At any gathering of n people must have the same hair colour. Step 1, if there is one person at the gathering then everyone at the gathering has the same hair colour. Step 2, Assume that k gathering of people must have the same hair colour. Step 3,…
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Prove that $3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all values of $n$

I have tried to prove this through mathematical induction but I can't seem to prove that the proposition works for $k+1$.
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Prove $n+2(n-1)+3(n-2)+...+2(n-1)+n={n(n+1)(n+2)\over 6}$ by induction

To prove: $n+2(n-1)+3(n-2)+...+2(n-1)+n={n(n+1)(n+2)\over 6}$ I assume the statement was derived from: $n+2(n-1)+3(n-2)+...+(n-1)(n-(n-2))+n(n-(n-1))={n(n+1)(n+2)\over 6}$ For $n=k+1$ $(k+1)+2(k)+3(k-1)+...+2(k)+(k+1)={(k+1)(k+2)(k+3)\over…
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What does induction show?

If I have a statement $P(n)$, and I show that $P(1)$ is true and $P(k)$ being true implies $P(k+1)$ is true, then it is my understanding that induction shows that $P(n)$ is true for all natural numbers. Suppose I have the statement $P(n)$ being…
user694996
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induction: 6 divides $(a^{2n+1}-a)$

Proof by induction: $\forall n \in \mathbb{N}$: 6 divides $(a^{2n+1}-a)$, $a \in \mathbb{N}$ Induction: for $n=1:$ $(a^{3}-a) = a*(a^{2}-1)=???$ How do I transform the expression further to get to the result.
franz3
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Question about induction for proof of Ramsey's Theorem

The induction step in this proof of Ramsey's theorem states that First we prove the theorem for the $2$-colour case, by induction on $r + s$. It is clear from the definition that for all $n$, $R(n, 1) = R(1, n) = 1$. This starts the induction. We…
Vasting
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$\forall n \in \mathbb{N}, 2 \leq n, \exists k \in \mathbb{N}$ such that $ 2^{k-1} \leq n \leq 2^{k}$

I am confused on how to proceed on the Inductive step for this problem. (sorry in advance if my Latex is trash, im still learning) I.H: Assume that $P(2), P(3), ...... P(m)$ holds where $m \in \mathbb{N}$ I broke it down into two cases. case $1$: $m…
user704972
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How to prove 1/(1*2)+1/(2*3) + … + 1/((2n-1)(2n)) = 1/(n+1) + 1/(n+2) + … + 1/(2n) by induction?

Basically, the expression in a task looks like this: I've made a thesis Then I use an assumption, to change some part of thesis: I replaced the red part with a right side of an assumption Then I substracted parts, which was the same in both sides…
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A question about the nuances of mathematical induction

Say I am trying to prove the statement $$1\cdot1! + 2\cdot 2! + \cdots + n \cdot n! = (n+1)!-1$$ by mathematical induction. I start by declaring that $P(n)$ is the statement given by $$1\cdot1! + 2\cdot 2! + \cdots + n \cdot n! = (n+1)!-1.$$ Now, I…
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Proof by induction: conversion to equal expressions

Lets suppose that I am trying to prove: $$1^2+...+n^2 = 1/6n(n+1)(2n+1)$$ via induction, but I am unable to convert (for sake of argument) $$1/6k(k+1)(2k+1)+(k+1)^2$$ to $$1/6(k+1)(k+2)(2k+3)$$ to prove that $$1^2+...+k^2+(k+1)^2 = 1/6k(k+1)(2k+1) +…