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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Proving terms are rational using Mathematical Induction

I was able to do the first part of the question, in the second part (Proof by Induction), I showed it holds for $n=1$: Then I Assumed its true for $n=k$. $$\begin{align}I_{k+1}=\frac{e^2}{2}-\frac{(k+1)}{2}I_k\end{align}$$ Then for…
M.S.E
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Why induction cannot be used for infinite sets?

Explain why induction cannot be used to conclude: $$\left(\bigcup_{n=1}^\infty A_n\right)^c = \bigcap_{n=1}^\infty A_n^c$$
Edie
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Vacuous truth and (simple and complete) induction

The way I understand complete induction, as applied to the naturals at least, the inductive step consists of assuming that a given proposition $p_i$ is true for $1 \le i \le n$, and from this deduce the truth of of $p_{n+1}$. However, I had thought…
wmnorth
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$6^{(n+2)} + 7^{(2n+1)}$ is divisible by $43$ for $n \ge 1$

Use mathematical induction to prove that 6(n+2) + 7(2n+1) is divisible by 43 for n >= 1. So start with n = 1: 6(1+2) + 7(2(1)+1) = 63 + 73 = 559 -> 559/43 = 13. So n=1 is divisible Let P(k): 6(k+2)+7(2k+1) , where k>=1 Show that P(k+1): 6((k+1)+2) +…
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Prove $n^2(n^4-1)$ is divisible by 60 using Mathematical Induction.

Base step: p(2)=4 * 15= 60 Inductive Hypothesis: Assuming p(k) = $k^2(k^4-1)$ = 60q Induction: p(k+1)= $(k+1)^2[(k+1)^4-1]$ = $(k+1)^2[(k+1)^2 + 1][(k+1)^2 - 1]$ = $(k+1)^2(k^2+2k+2)(k^2+2k+1- 1)$ = $(k+1)^2(k^2+2k+2) k (k+2)$ …
Niharika
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$19 \mid 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}}$

I tried to demonstrate the next equation is divisible by 19: $$ 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}} $$ When $n$ is $1$: $$ 2^{2^1} + 3^{2^1} + 5^{2^1} $$ $$ 4 + 9 + 25 = 38 $$ When $n$ is $k$: $$ 2^{2^k} + 3^{2^k} + 5^{2^k} $$ Finally, when $n$ is…
jtwalters
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Mathematical induction--When it can and can't be used

I'm working through a problem set on mathematical induction. One of the problems asks you to prove that for all $n\in\mathbb N$, $$\sum_{i=0}^{n}8n-5=4n^2-n.$$ I don't have a problem proving this, but I wondered: What formula relates the…
Ryan
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Prove that $\log(x) < x$ for $x > 0$, $x\in \mathbb{N}$.

I'm trying to prove $ \log(x) < x$ for $x > 0$ by induction. Base case: $x = 1$ $\log (1) < 1$ ---> $0 < 1$ which is certainly true. Inductive hypothesis: Assume $x = k$ ---> $\log(k) < k$ for $k > 0$ Inductive conclusion: Prove $\log(k+1) <…
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Prove GM-AM inequality using induction

Show that $G_{2^n}\le A_{2^n}$ by using induction on n. I've proven the base case in the previous exercise: Let $G_2=\sqrt{a_1a_2}$ and $A_2=\frac{1}{2}(a_1+a_2)$ and $a_1,a_2 \in \mathbb{R}$ $$\sqrt{a_1a_2}\le…
russCam
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Induction Question: $7^n-6n-1=36m$

In the end I get the following: $$(7^n-6k-1)+6(7^k-1)$$ I tell myself that $(7^n-6k-1)$ is clearly divisible by $36$ by the initial proposition, and that $6(7^k-1)$ is divisible by $36$ because $7^k-1$ is divisible by $6$ by the initial…
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Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $

Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $. I provide the answer below.
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Prove reversal of a string by induction

I am trying to prove that: (uv)R = vRuR where R is the reversal of a String defined recursively as: aR = a (wa)R = awR I think I have the base case right, but I am having trouble with the inductive step and final proof. here is what I have: Base…
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Using the method of induction to show

How can I use the method of induction to show for any real number $r$ does not equal $1$ and any positive integer $n$ show that $$1+r+r^2+\cdots+r^n=\frac{1r^{n+1}-1}{r-1}$$ for $n=1$ it seems to work $$1+r+\cdots+r^n=(1+r)$$ then…
Fernando Martinez
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How to proceed to show this hold by induction?

Show that $$\frac3{1\cdot2\cdot4}+\frac4{2\cdot3\cdot5}+\dots+\frac{n+2}{n(n+1)(n+3)}=\frac16\left[\frac{29}6-\frac4{n+1}-\frac1{n+2}-\frac1{n+3}\right],\text{ for $n\in\mathbb N$}.$$ I try induction. For $n=1$, it is trivial and then let it is…
Silent
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Proof: $2^{n-1}(a^n+b^n)>(a+b)^n$

If $n \in \mathbb{N}$ with $n \geq 2$ and $a,b \in \mathbb{R}$ with $a+b >0$ and $a \neq b$, then $$2^{n-1}(a^n+b^n)>(a+b)^n.$$ I tried to do it with induction. The induction basis was no problem but I got stuck in the induction step: $n \to…