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 $N\in\mathbb N$.

10150 questions
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Proofing Leibnitz formula ($(uv)^{k} = \sum_{m=0}^{k}\binom{k}{m}u^m v^{k-m}$) using mathematical induction

I am currently trying to learn advanced mathematics. So I'm currently reading 'The Art of Proof' by Matthias Beck and Ross Geoghegan. I tried to solve one of the problems but I don't know if I'm actually close to an answer or not. The thing is I'm…
Parousia
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Ordinary Induction vs. Strong Induction (on the same example)

I'm looking for a solid explanation to grow in intuition for strong induction after having a decent foundation in ordinary induction. For example, The proof by induction for "Prove by induction for each $n \in \mathbb{N}$, $3$ divides $4^n$" is as…
kr1tzy
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Prove by induction: $(n-1)!>2^n$ for all $n \geq 6$

Let $P(n)$ be the statement $(n-1)!>2^n$ Use mathematical induction to show that $P(n)$ is true for all $n \geq 6$. I've got $(k)!>2(k-1)!$. With the given value of $k= 6$ is this proof enough?
RM1973
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Difference between downward induction and infinite descent

The principle of infinite descent states that if the truth of $P(n)$ for some natural number $n$ implies the truth of $P(m)$ for some natural number $m
Maxis Jaisi
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how: mathematical induction prove inequation

Provided that $p\geq-1$, prove $(1+p)^n\geq1+np$ for all integers $n\geq 0$ Also, where in the calculation do I use $p\geq -1$? Thanks guy!
Maria
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Use induction to prove that the inequality, $n! < \left(\frac{n}{2}\right)^n$ continues to hold after $n=6$.

I am confused how you would continue with the inductive step. Thanks
JanoyCresva
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I have done this induction proof for the Fibonacci-sequence

Would you please let me know how do I improve this proof? The numbers $$F_{0},F_{1},F_{2},...$$ are defined as follows (this is a definition by mathematical induction, by the way): $$F_{0} = 0, F_{1} = 1, F_{n+2} = F_{n+1} + F_{n} \ for \…
DSL
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Using mathematical induction prove that $ 11 \cdot 3^n + 3 \cdot 7^n - 6$ is divisible by 8

Prove that $ \phi(n) =11 \cdot 3^n + 3 \cdot 7^n - 6 $ is divisible by 8 for all $n \in N$. Base: $ n = 0 $ $ 8 | 11 + 3 - 6 $ is obvious. Now let $\phi(n)$ be true we now prove that is also true for $ \phi(n+1)$. So we get $ 11 \cdot 3^{n+1} + 3…
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Proof of property by mathematical Induction

I need help executing a mathematical Induction for the following equation: add(x,succ(y)) = succ(add(x,y)) I know that I have to either substitute n or m for "zero" for the Base Case, but I don't know what exactly I have to do after this. To show:…
hey hoe
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possible bills one can create with two types of bills

My friend asked me this question, and it's been bugging my mind There are 5€ bills and 8€ bills, what are all the possible bills one can create with these two types of bills. Prove your answer the only hint is to use strong induction How can I…
swhat7
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Prove by Induction that $n! > \left(\frac{n}{e}\right)^n$ for $n>1$

I'm studying a combinatorics text (Cameron) in preparation for taking discrete math this upcoming semester. I've taken a fair amount of math in college, through Calculus II, but this is my first 500+ level math course. Here is the problem that I'm…
quantif
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Is this a legit form of induction?

I'm currently studying a proof in graph theory and it says: 'Suppose that what we want to prove is true for 1,2,3, ..., n-1. We will prove that it holds for n'. Is this a legit form of induction? Usually, in the induction proofs I encountered until…
user370967
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Prove using induction that $2^{4^n}+5$ is divisible by 21

I have to show, using induction, that $2^{4^n}+5$ is divisible by $21$. It is supposed to be a standard exercise, but no matter what I try, I get to a point where I have to use two more inductions. For example, here is one of the things I…
Ludolila
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Induction: $\sqrt{1+\sqrt{n+\sqrt{n^2+\sqrt{n^3+...+\sqrt{n^n}}}}} < n$

Prove using induction for all natural $n>1$. I can't seem to find a way to use the assumption (for $n$) to prove the $n+1$ case. Thanks ahead.
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Induction proof of inequality

the problem is: Prove by induction. In each case, n is a positive integer. $2^{n} ≤ 2^{(n+1)} - 2^{(n-1)} - 1$ Right now I am at the following, but I got stuck and don't know where to go from here. $2^{(n+1)} ≤ 2^{(n+2)} - 2^{n} - 1 =…