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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Explaining Proofs by Induction

We consider the statements $P (n)$: "$4^n −1$ is divisible by $3$" and $Q(n)$: "$4^n + 1$ is divisible by $3$" where $n \in \mathbb{N}$. (1) Show that both inductive steps $P (n) \implies P (n + 1)$ and $Q(n) \implies Q(n + 1)$ are true for all $n…
Michelle Drolet
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Proof for Total Overhead Formula for IPv4 Fragmentation

I am struggling to develop a proof for an equation to compute the total bandwidth required for fragmented IP packets. When a router forwards an IPv4 packet of size $n$ over a link with an MTU $m$ then the packet will be fragmented if $n>m$. Each…
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Sum of first $1+3+9+\cdots+3^n$ natural numbers.

How to prove that the sum of the first $1+3+9+\cdots+3^n$ natural numbers is equal to $1^2+3^2+9^2+\cdots+(3^n)^2$? I've tried induction, but I can't get through the induction step. The base is simple, but in the step I can only use the induction…
implicati0n
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Proof by strong induction of $(a + b\sqrt{2})^n = a_n + b_n \sqrt{2}$

I'm not very familiar with proofs by strong induction. I have a sketch for this one but Iot quite sure about it's validaty. Let $a$, $b$, $a_n$, $b_n$ be integers such that. $$(a + b\sqrt{2})^n = a_n + b_n \sqrt{2}$$ where $a$ is the integer closest…
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proof by induction of $n^n\le(n!)^2\ \ \ \forall n\in \mathbb{N}^+$

Prove by induction that: $$n^n \le (n!)^2\ \ , \ \forall n\in \mathbb{N}^+$$ my attempt was: $$(n+1)^{n-1} \le n^n \le (n!)^2\ \ , \ \forall n\in \mathbb{N}^+$$ which comes down to $$\ \ \ \ \ (n+1)^{n-1} \le n^n \ , \ \forall n\in…
David
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Prove via induction $\frac{(n+1)^2}{2^n}\le\frac{9}{4}$

How do you prove in induction that: $$\frac{(n+1)^2}{2^n}\le\frac{9}{4}$$ This is what I keep getting: Checking for $n=1$ we get $2\le\frac{9}{4}$. Assuming it's true for $n$ and checking for $n+1$ I get…
yuvalz
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Prove, by mathematical induction, a confusing answer

The question is to prove (or disprove) by mathematical induction the following statement: For $ n, a, r, \in \mathbb Z,$ and $n, a, r \in Z_{>0},$ and $r ≥ 2$ $$S(n) = \sum_{i=0}^n ar^{i-1} = an$$ The sample answer given by the book is…
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Help with summation induction

Prove that $$\sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i } } = } 2-\frac { n+2 }{ { 2 }^{ n } } \quad { for\quad all}\quad n\in \mathbb N $$ My attempt: For my inductive step I tried the following: $$\left(2 − \left(\frac{k +…
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Define $({a}_{n})$ inductively by putting ${a}_{1}={a}_{2}=1$ and ${a}_{n+2}={a}_{n}+{a}_{n+1}$ $\forall n\in N$. prove that $\lim {x}_{n}=\phi$

Define $({a}_{n})$ inductively by putting ${a}_{1}={a}_{2}=1$ and ${a}_{n+2}={a}_{n}+{a}_{n+1}$, $\forall n\in N$. Write ${x}_{n}=\frac{{a}_{n}}{{a}_{n+1}}$ and prove that $\lim {x}_{n}=\phi$, where $\phi$ is the only positive real such that…
Shatrix
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Prove $5^n +5 <5^{n+1}$ $∀n∈N$ using induction

Prove $5^n +5 <5^{n+1}$ $∀n∈N$ Base Case: $n=1$ $\implies 5^1 +5 <25$ $\implies 10<25$ ; holds true Induction hypothesis: Suppose $5^k +5 < 5^{k+1}$ is true for k∈N Then; $\implies 5^{k+1} +5 < 5^{k+2}$ $\implies 5\cdot 5^k +5 < 25*5^k$ I don't…
TheGamer
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Proof by mathematical induction, difficulties

Usually, proofs by mathematical induction are not that difficult (at least the ones I encountered so far in my mathematical journey), but I am stuck with this one... Spent at least 2-3 hours on it. I just can't see it... $$ \sum_{i=0}^n…
Shocky2
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Let $f$ be a real function such that, for $x,y\in \mathbb{R}$ $f(x+y)=f(x)+f(y)$ Prove that (a) $f(0)=0$ (b) $f(n)=nf(1)$

Let $f$ be a real function such that, for $x,y\in \mathbb{R}$ $$f(x+y)=f(x)+f(y)$$ Prove that (a) $f(0)=0$ (b) $f(n)=nf(1)$ I have no clue how to prove this, would anyone give me some primer to start with?
DSL
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Lack of understanding of steps for second principle of finite induction for: $a^n -1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-3}+\cdots+a+1)$

The problem i'm trying to solve is: Use the second principle of finite induction to establish that: $$a^n -1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-3}+\cdots+a+1)$$ for all $n \ge 1$ I have found a solution online that reads: $$ \mbox{For }k=1: \ a^1-1…
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Simple proof by induction for $a+ar+ar^2+...+ar^n$

If $r\neq 1$, show that $$a+ar+ar^2+\cdots+ar^n=\frac{a\left(r^{n+1}-1\right)}{r-1}$$ for any positive integer $n$ I seem to be doing something wrong could somebody tell me what is wrong with my…
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Why do we mention that well-ordering principle implies there is a least element?

It seems to me that the purpose of the well-ordering principle is to prove that something that is a natural number exists and proving that something exists really has nothing to do with having a least element.. unless of course we must specify what…
user420360