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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2^(x-1) + 2^(x-2) + 2^(x-3) < 2^x

Consider the recurrence: tn ={ \begin{array}{ll} 1 & n = 1 \\ 1 & n = 2 \\ 1 & n = 3 \\ t_{n-1} + t_{n-2} + t_{n-3} & n \geq 4 \end{array} For all natural numbers $n$, starting at $1$, $t_n < 2^n$
Mus
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The set RAF of rational functions of one real variable is the set of functions defined recursively as follows:

I have no clue where to start on this problem... Here is the full problem (couldn't fit it on top). The set RAF of rational functions of one real variable is the set of functions defined recursively as follows: Base cases: ✏ The identity function,…
Nate123
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induction inequality

I'm trying to solve the induction: $ 1 + \frac{1}{\sqrt2} + ...+ \frac{1}{\sqrt{n}} > 2\sqrt{n+1} -2 $ My thought for solving this was: $f(k) = 1 + \frac{1}{\sqrt2} + ...+ \frac{1}{\sqrt{k}}$ $f(k) + \frac{1}{\sqrt{k+1}} > 2\sqrt{k+2} -2$ - Need to…
Moran Tailu
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Proof by Induction - $(\sum_{i=1}^{n}i)^2 = \sum_{i=1}^{n}i^3$

Hello Mathematics Community, currently I am trying to prove: $(\sum_{i=1}^{n+1}i)^2 = \sum_{i=1}^{n}i^3$ by Induction. My Progress: Assumption: $(\sum_{i=1}^{n+1}i)^2 = \sum_{i=1}^{n}i^3$ Induction Hypothesis: $=(\sum_{i=1}^{n+1}i)^2 =…
M.Hisoka
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Strong Induction with an inequality.

For all integers $n\geq 1$, $$\frac{1}{2}\cdot \frac{3}{4} \cdot\cdots\cdot \frac{2n-1}{2n} \leq \frac{1}{\sqrt{3n + 1}}.$$ Need help with inductive step. IB: P(1) $$\prod_{i=1}^{1} \frac{2i-1}{2i} = \frac{1}{2} \leq …
Arualia
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Proof by induction with a variable

I need to prove by induction that for every $n \ge 1$, $n\in\mathbb{N}$ and $a>0$ the following statement holds $$(1+a)^n \ge 1 + na + \dfrac{n(n-1)}{2}a^2$$ The statement is true for $n=1$: $$1+a \ge 1+a$$ Assume for some $n=k$: $$(1+a)^k \ge 1 +…
McLovin
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$f(n) \leq cn^2$ find a positive integer c

Base Case: Let n = 0 $f(n) = 4$ [def of f] and $cn^2 = 25c$ Therefore we need $4 \leq 25c$ or $c \geq \frac{4}{25}$ (*) IND STEP: Let $n > 0$. Suppose $f(n) \leq cn^2$ [IH] Trying to show $f(n+1) \leq c(n+1)^2$ $f(n+1) = 7f\left(\lfloor…
Tinler
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for all $n \in \mathbb N f(n) \leq 7 \cdot 5^n - 3^n$, also $f(n) \leq c \cdot 5^n$

Base Case: Let n = 0 $f(n) = 6$ [def of f] $\leq 7 - 1 = 6 = 7 \cdot 5^0 - 3^0 = 7 \cdot 5^n - 3^n$ as wanted IND STEP: Let $n > 0$. Suppose $f(n) \leq 7 \cdot 5^n - 3^n$ [IH] Trying to show: $f(n+1) \leq 7 \cdot 5^{n+1} - 3^{n+1}$ $f(n+1) = 5f(n)…
Tinler
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Recurrence induction finding positive integer b such that $f(n) \leq bn$

Use induction to find an integer $b > 0$ such that $f(n) \leq bn$ for all $n > 0$ My Attempt! Basis: Let $n = 1$ $f(n) = 6$ [Def of f] and $bn = b$ Therefore we need $b \geq 6$ Let $n = 2$ $f(n) = 12$ [Def of f] and $bn = 2b$ Therefore we need $b…
Tinler
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Two questions about proof by induction

Let $P(n)$ be a statement to be proven by induction, where $n \in \mathbb{N}$ (with possible exceptions). In the inductive step we conventionally assume a statement $P(k)$ to be true, to show that $P(k) \implies P(k+1)$. Do we have to define $k$ as…
Daphne
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How to proof that for all n $ (\sum_{i=1}^n i)^2 = \sum_{i=1}^n i^3 $

What i had so far was this: claim: How to proof that for all n $$ \left(\sum_{i=1}^n i\right)^2 = \sum_{i=1}^n i^3 $$ Proof To proof: How to proof that for all n $$ \left(\sum_{i=1}^n i\right)^2 = \sum_{i=1}^n i^3 $$ basis $n=1$: $$…
bassie
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Fibonacci sequence - an inductive proof of its formula

Prove that $$a_n = \frac{1}{\sqrt{5}}\left((\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n\right)$$ if $a_{n+1} = a_{n-1} + a_n$ My idea to solve this was to do what follows: (1) Check if this relation holds for $n =1$ and $n=2$ (2) Assume…
Aemilius
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induction method- an interesting application

In a pyramid scheme, the members of a company recruit more members into the company. A person that has already been recruited cannot be recruited again. We will also assume that no one leaves the company once they are recruited.A person's level in…
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Prove Geometric Progression By Mathematical Induction

Can someone please check this for me? Like the wordings and working. Thanks The question is: Use Mathematical Induction to prove that $$\sum_{i=1}^{n}i2^i=(n-1)2^{n+1}+2.$$ My answer: $$\sum_{i=1}^{n}i2^i=(n-1)2^{n+1}+2.$$ My answer: let…
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Question about Induction principle, its usage and general proof.

There were weak induction and strong induction. Question (1) When can we use weak induction or strong induction? Weak induction seemed to use the nature number as it was assuming $n$ to $n+1$, but we can say all the cases before $\alpha$ was true in…
user416486