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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Can't do this induction proof with sqrt(4) repeating

Ok so I realize that the last term will be sqrt(6) but I just don't know how to manipulate this expressions to make it provable by induction. I tried rewriting it using exponents but had no luck with that
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Recursive definition of multiplication

I have the following function: $$ \begin{cases} mul (a, 0) = 0&\mbox{if }n=0\\ mul (a, n) = mul (2a, \frac{n}{2})&\mbox{if }n\mbox{ is even}\\ mul (a, n) = mul (2a, \frac{n-1}{2}) + a&\mbox{if }n\mbox{ is odd}\\ \end{cases} $$ It is clear that the…
Jose
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Prove that for $ \displaystyle n≥1, \sum_ {i=1} ^{ n} \dfrac 1 {i^2} ≤ 2 - \dfrac 1 n$

Can someone help me with this proof? Prove that for $n≥1$, $\displaystyle\sum_{i=1}^n \dfrac 1 {i^2} \le 2 -\dfrac 1 n$. The Base Case is easy but for my induction step, I get stuck at... (sum from i=1 to k+1 of 1/i^2) ≤ 2 - 1/(k+1) (sum…
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Proof of equation $\sum_{k}{n\brack k}a_k = n!2^{n-1}$ by induction

I'm trying to prove to following equation: $$\sum_{k=0}^{n}{n\brack k} a_k = n!2^{n-1};\ \ \ n\ge 1$$ $a_n$ - number of ordered partition of set. We have following recursion dependencies: $a_n = \sum_{k=0}^{n-1}{n\choose k}a_k;\ \ \ \ a_0 = 1…
xawey
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Mathematical Induction proof for $(n!)^2 > n^n$

I have the math problem (induction proof - $n!^2 > n^n$) that I try to solve and I haven't yet managed to get it right so maybe somebody could help me. My current plan solving the problem is the following : 1) for : $n = 1 \implies$ False; $n = 2…
Mihai
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Proof by induction that $\sum_{i=1}^{n}(3i-2)=\frac{n}{2}(3n-1)$

I am trying to perform this proof but I find myself stuck Prove for all natural number n. $\sum_{i=1}^{n}(3i-2)=\frac{n}{2}(3n-1)$ The first step ofcourse is P(1) because 1 is the first natural number. $1=3-2$ Second step let $n=k$ the ass umption…
Fernando Martinez
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induction proof of recursive multiplication

mul(a,0) = 0 mul(a,n) = if a%2 then mul(2a,n/2) else mul(2a, (n-1)/2)+a mul(a,n) = a*n
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How is derived the inductive step in mathematical induction?

I am quite familiar with the algorithm of mathematical induction but I can't rationalize the inductive step very well. Suppose I have the classical example: $$0 + 1 +2 + \ldots + n = \frac{n(n+1)}{2}$$ In the inductive step I have to show that this…
G M
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Induction with negative step

We've learned that we can use induction to show that a statement holds for all natural numbers (or for all natural numbers above n). The steps are: prove that the statement holds for a base number b assuming that the statement holds for n, show…
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Proving by induction: $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $

WTS $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ for all natural $n$. Have checked $P_1$, and assumed $P_k$. Trying the following argument: $P_{k+1} = \frac{1\cdot3\cdot5\cdot \ldots \cdot…
Alec
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A question on mathematical Induction: $f(a)=f(a^2)$ implies $f(a)=f(a^{2^n})$

I came across this question in an answer given to a question here on MSE. Given that $f(a)=f(a^2)$ for all $a\in [0,1]$,$f$ continuous, it is easy to prove by induction that $p(n): f(a) = f(a^{2^{n}})$ for all $a\in [0,1]$ and for all $n\in…
Jon
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Deduce that the next integer greater

Deduce that the next integer greater than $(3+\sqrt 5)^n$ is divisible by $2^n$ I tried expanding it by binomial theorem but got nothing
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odd/even binomial coefficient identity

For all n\geq1 : $$\left(\begin{matrix}2n\\ 0 \end{matrix}\right) +\left(\begin{matrix}2n\\ 2 \end{matrix}\right) +\left(\begin{matrix}2n\\ 4 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2k \end{matrix}\right) + \ldots …
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Induction: the complement A1 U A2 … U An is the intersection of Ac 1, Ac 2, …, Ac n

Prove by induction that the complement of $ A1 \cup A2...An = A1^c \cap A2^c ...\cap An^c$ My approach: basic step is true, $\overline A1 = A1^c$, then assume $ A1 \cup A2...Ak = A1^c \cap A2^c ...\cap Ak^c$, prove the case of $k+1$ is true. How…
Gavin Z.
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Proof for $\sqrt[n]{n!}\le{{n+1}\over2}$ using induction

Prove $\sqrt[n]{n!}\le{{n+1}\over2}$, $n \in N^+$, using induction. This is how far I got, but then I got stuck: $$\sqrt[n+1]{(n+1)!}\le{{n+2}\over2}$$ $$((n+1)!)^{1\over{n+1}}\le{n\over2}+1$$ $$((n+1)\times…
E.O.
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