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: $\frac{(2n)!}{n!(n+1)!}\in \mathbb{N}, \forall n\in\mathbb{N}$

Prove: $\frac{(2n)!}{n!(n+1)!}\in \mathbb{N}, \forall n\in\mathbb{N}$ I used induction, for $n=k$ assume that $\frac{(2k)!}{k!(k+1)!}\in \mathbb{N}$ For $n=k+1$ $\frac{(2k+2)!}{(k+1)!(k+2)!}\in \mathbb{N}$ $x=\frac{(2k+2)!}{(k+1)!(k+2)!}\cdot…
user300045
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Induction from n to 1

I have an algorithm with a for loop that goes from i=n-1 to 1. I'm attempting to prove it's correctness. Am I allowed to use induction like so step 1 -- base case i = n step 2 -- induction hypothesis, assume it works for i = n + 1 step 3 -- prove…
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What is wrong with the proof

What is wrong with the following “proof” by Mathematical Induction? We will prove statements that all computers are built by the same manufacturer. In particular, we prove statements B(n) with n ∈ N, that “in any collection of n computers, all of…
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Algebraic Manipulation for Mathematical Induction

I'm working on a mathematical induction problem. The question is as follows: $P = \begin{pmatrix} 1-A & A \\ B & 1-B \\ \end{pmatrix}$ for A,B $\epsilon$ (0, 1). Show by induction, or otherwise that $P^n = \frac{1}{A+B} \begin{pmatrix} B & A \\ B &…
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How to prove this inductive form?

Assuming $\exists n \ge 2: P_n$, $\;\neg P_k \implies \neg P_{k-1}$, and $\neg P_1$, is it valid to conclude $\exists! k , 1 < k \le n: P_k \wedge \neg P_{k-1}$? What theorem or technique would I need in order to prove this?
AJMansfield
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How can I develop this using induction?

I'm trying to prove using induction that $a^n +b^n = (a-b) \cdot \sum\limits_{k=1}^{n}a^{n-k}b^k-1$ So I have developed the expression for $n+1$ but I get to $a \cdot b^n - b^{n+1} + a^n - b^n$ And from here I can't simplify until $a^{n+1} + b^{n+1}…
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Prove by Induction help?

Prove by induction that for $1 \le n$: $$\sum_{k=1}^n k(k + 1)(k + 2) = 6 + 24 + . . . + n(n + 1)(n + 2) = \frac 14 n(n + 1)(n + 2)(n + 3) $$ Basis step: I got $n(1) = 6$ and $n(2) = 30 \ (24 + 6 = 30)$ Assumption: $n =…
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Proof by Induction Using Fibonacci -- Not Sure About Other Question's Answers

Here was my previous question: Proof by Induction Using Fibonacci numbers There was a similar one in existence already over here: Inductive proof of a formula for Fibonacci numbers The one that was answered already had a decent answer but it isn't…
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A proof by strong induction for a recurrences defining function

Problem: Consider the following recurrences defining function, $$f_2(n) = \lbrace 1\ if\ n = 0;\ 1 + \sum_{i=0}^{n-1} f_2(i)\ if\ n > 0\rbrace, f_2: N \to N $$ Prove $f_2(n) = 2^n$ My attempt: Proof by strong induction. so, Base Cases: $$n = 0…
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Induction proof of sentence

How do I prove that $\prod\limits_{i=1}^{n} (\frac{2i - 1}{2i}) \le \frac{1}{\sqrt{3n + 1}} $ using induction for n > 0 ? In general, for (n+1) case, I split the productory in: factor(n+1) * productory [i, n] factor(i) given productory [i, n]…
DcCoO
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the sequence of Fibonacci numbers is defined as follows: $x_1=1, x_2=1$

the sequence of Fibonacci numbers is defined as follows: $x_1=1, x_2=1$, and, for $n>2, x_n=x_{n-1} + x_{n-2}$. Prove that $$ x_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right )^n - \left (\frac{1-\sqrt{5}}{2}\right )^n\right ] $$ for…
Iren
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Trouble with induction on polynomials

Use induction to prove the following: $1a)$ Show for all positive integers $n$ that $$ \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $$ $1b)$ Show that if $p$ and $q$ are polynomials so that $i)$ $q(0)=0$ $ii)$ $p(n)=q(n)-q(n-1)$ for all $n$…
kvasir
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Proof by induction the following: $n+1 ≤ 2^n ≤ (n+1)!$

I've done the basis step for $n=1$ and managed to arrange the $n=k+1$ to: $(k+1) + 1 ≤ 2\cdot2^k ≤ (k+1)!(k+2)$ Not sure how to proceed from here?
Mals T
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Simple induction when dealing with floor

NOTE: This MUST be done with simple induction I'm currently stuck on properly proving induction on a set when floor is involved. I have made up a question to illustrate the issue, it may not be a great question but hopefully you can see where I'm…
Water
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$P(0), P(1)$ hold and $P(n) → P(n + 2)$ for $n\geq 1$. For which $n$ is $P(n)$ T?

The question is: $P(0)$ hold $P(1)$ hold $P(n) \rightarrow P(n + 2)$ for $n \geq 1$ For which values of $n$ does $P(n)$ hold? My initial answer was that $P(n)$ holds for all odd positive values of $n$, but now I am not so sure. The reason I…