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 by induction $u_{n+2}+u_n=4u_{n+1}$ for all $n\in \mathbb{N}$, where $u_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n$.

Prove by induction $u_{n+2}+u_n=4u_{n+1}$ for all $n\in \mathbb{N}$, where $u_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n$. For $n=1$, My efforts: LHS= $u_3+u_1=(2+\sqrt{3})^3+(2-\sqrt{3})^3+ (2+\sqrt{3})^1+(2-\sqrt{3})^1=56$. RHS=…
BSFU
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$P(n) = 2^n > n^2$, find $k \in \mathbb{N}$ so that $P(k) \Rightarrow P(k+1)$

I've got this problem where I need to find $k \in \mathbb{N}$ so that $P(k) \Rightarrow P(k+1)$ and $P(n) = 2^n \gt n^2$ By induction I have: $2 = 2^1 > 1^2 = 1$ which is ok with the first condition. Now I'm having troubles with the next step: If…
Lucas
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Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of $3$

Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck. Show that $1+2^n+2^{2n}$ is divisible by $7$,…
Rob
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Induction Proof for Extended sum/product rule | Combinatorics

So basically I have to prove by induction that: i) Given finite sets $A_{1},A_{2},...,A_{n}$ which are pairwise disjoint, then $|\cup_{i = 1}^{n}A_{i} | = \sum_{i = 1}^{n}|A_{i}|$. ii) Given finite sets $A_{1},A_{2},...,A_{n}$, then $|\prod_{i =…
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Prove by induction: $\dfrac{d^{2n}}{dx^{2n}}(x^2-1)^n = (2n)!$

Let $P_n$ be the statement that $\dfrac{d^{2n}}{dx^{2n}}(x^2-1)^n = (2n)!$ Base case: n = 0, $\dfrac{d^0}{dx^0}(x^2-1)^0 = 1 = 0!$ Assume $P_m = \dfrac{d^m}{dx^m}(x^2-1)^m = m!$ is true. Prove $P_{m+1} =…
Joe
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The argument of a proof based on double induction

I am struggling to convince myself of this proof. Let me rewrite it so that the proof's structure and my interpretation of it are more apparent. Let $ S(k, n) $ be true when $ n! \mid P(k, n) $ where $ P(k, n) = (k+1)(k+2)\cdots(k+n) $. We want to…
x63
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The result of the bi-conditional $\leftrightarrow$ is true when n is odd and m is odd, and when n is even and m is even. Proof by induction

Let m be the number of Truth values for each row, and let n be the size of the truth table. $2^n$. I discovered that the result of the bi-conditional $p1\leftrightarrow p2$ $\leftrightarrow p3$ is true when n is odd and m is odd. I've also…
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Prove $p_1 \iff p_2 \iff p_3 \iff \ldots \iff p_n$ by using Proof by Induction.

I had to write a Python program that shows the proper output of an "If and only If" truth table. My program works, and now I have to use induction that my formula is correct for all positive integers. We studied proof by induction a few weeks ago,…
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Inductive hypothesis of constructive induction

Let $\Sigma$ be an alphabet and $\Sigma^*$ its Kleene closure (the set of all possible finite strings over the alphabet). Also, define the set of prefixes of $x \in \Sigma^*$ by $$ \mathrm{pre}(x) = \{x[1:i] \mid i \in \{1, \ldots, |x|\}\}, $$ where…
sesodesa
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skipping proof through induction

My professor showed following problem and he solved through induction. However, I'm not sure why this problem can't be solved the way I did it, since every step seems to be true. For all natural numbers $n$ following holds true: $$…
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How can I prove that I am using valid mathematical induction?

I am retaking a discrete math class and my new teacher is teaching mathematical induction differently than my old teacher. I really like the way I first learned it because I think it is more formal and I understand it better. But my current teacher…
Ryan
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Prove that the number of characters in a language is even, by structural induction

If we have a language $L$ defined over the alphabet ${(,)}$ defined by $\epsilon \in L$ If $X \in L$ and $Y \in L$ then $XY \in L$ If $X \in L$ then $(X) \in L$ I want to prove that there is an even number of characters in any $X \in L$. The…
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Prove $n - 2 < \frac{n^2 - n}{12}$ by Mathematical Induction

I am trying to prove the following $n - 2 < (n^2 - n)/12$ when $n > 10$ by Mathematical Induction. The following is what I've come up so far (please bare with me): Property to be proven $P(n)$: $$ n - 2 < (n^2 - n)/12 \hspace{.5cm}\leftarrow…
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proof by using induction

I'm trying to prove the following by induction but I'm stuck. $x_1 = 1, x_2 = 2, x_n=\frac{1}{2}(x_{n-1}+x_{n-2})$. Show that: $x_n-x_{n+1} = \frac{(-1)^n}{2^{n-1}}$ I proved the basis step, but I'm stuck in the inductive step. I tried going from…
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Proving recurrence using induction where there is an upper limit on the number of integers to prove it for

Given $x_0=1$ and $x_j=x_{j-1}\frac{N-(j-1)}{N}+x_{j+1}\frac{j+1}{N}$ for $j=1,...,N-1$, the formula $x_j={N\choose j}$ can be proven by induction. I do not see why we are able prove it by induction, seeing that using induction, we prove the base…
johnson
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