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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Proof of $(a+b)^{n+1}$

I have to do proof of $(a+b)^n$ and $(a+b)^{n+1}$ with mathematical induction. I finished the first one $(a+b)^n = a^n + na^{n-1}b+\dots+b^n$. I however have trouble with the second one, I don't know what to start exactly. Any help/hints?
user95523
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Odd-Even Mathematical Induction

Simpler version of the Josephus problem. Circle of $n$ people, every second is removed let $f(n)$ represent the last one from a circle of $n$ people I've proved $f(1)=1$; $f(2n)= 2f(n) - 1$; $f(2n+1) = 2f(n) + 1$ Hypothesis: $f(2^m + l)=1+2l$, $2^m…
Ve5ko_95
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Examples of statements that could be proven using strong induction without checking a special case?

After two days of contemplating, I believe that I've finally realized why strong induction doesn't require a base case as in regular induction. However, there's something I'd like to ask. To make it easier for discussion, let me restate the…
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More Bases for Strong Induction - Supersedes Weak Induction?

I recently learned about strong induction, and had a couple questions. First, on sites such as this one: http://www.mathblog.dk/strong-induction/ , it is said that using strong induction requires more bases case. (This makes sense to me, as you are…
user82004
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Proof by induction: $n^n\leq 3^nn!$ for any $n\geq1$

I am struggling a lot trying to prove by indiction this inequality: $$n^n\leq 3^nn! \text{ for any } n\geq1.$$ Could you please help me? When I try to prove it for $n+1$, I find myself in this condition: $$(n+1)^{n+1}\leq 3(n+1)3^nn!.$$ But I’m not…
Gaudio
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Prove by induction $2^n < n!$

Decide for which $n ∈ \mathbb N$ the inequality $2^n <2!$ is true and prove your claim by induction. If $ n = 1$ hence $ 2^1 \not<1!$ If $ n = 2$ hence $ 2^2 \not<2!$ If $ n = 3$ hence $…
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Mathematical induction problem: $\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{n(n+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$

I have problem solving this induction. I must be missing something because i don't know how to do it. Can someone help? Thanks $$\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{n(n+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$$
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Use mathematical induction to prove that $5^{2n+1}$ + $3^{2n+1}$ is divisible by $4$ for all natural numbers $n$.

Use mathematical induction to prove that $5^{2n+1}$ + $3^{2n+1}$ is divisible by $4$ for all natural numbers $n$. I hope I'm going in the right direction at least. Here is where I got to so far: When $n = 0$, we have $5^{2(0)+1} + 3^{2(0)+1} = 8$,…
i.diazr
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Can't find a reason for equation becoming negative in induction proof

I am currently stuck on an induction proof where I don't seem to understand why the positive 2(k + 1) becomes negative after giving both parts of the equation equal denominators. What is the reason for this? thank's a lot in advance! If it is…
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induction assuming $P(n)$ and $P(n - 1)$

So I understand that induction leads from the fifth axiom of natural numbers, where we say that we show the statement must be true for P(n + 1) if it is true for P(n) for some arbitrary n, then for all P(n) (starting at our base case), it must be…
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Discrete Math : proving of predicate logic

Assume $P(n)$ is a statement about integers $n\ge 1$ such that $∀n,P(n) \implies P (n + 4)$. Is the following statement $(∀n,P(n))$ always true, or always false, or can be true for some values and false for some other values?
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Proof by induction that $(n-1)^3 + (n+1)^3$ is divisible by 4 for all $n \in \mathbb{N}^+$

I want to prove that $(n-1)^3 + (n+1)^3$ is divisible by 4 for all $n \in \mathbb{N}^+$. I have looked up solutions for similar problems but cannot make it work for this one. The expression $(n-1)^3 + (n+1)^3$ can be simplified to $2n^3 + 6n$…
Mohi
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Question on induction

For which $n\geq0$ is $3^n>n^3$ valid? Consider the four first cases: $$3^0=1>0$$ $$3^1=3>1^3$$ $$3^2=9>2^3$$ $$3^3=27\ngtr3^3$$ So let us consider the base case $k=4$: $$3^4=81 >4^3,$$ which holds. Then we need to show $3^{k+1}>(k+1)^3…
Luthier415Hz
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Proof by induction $(1+\frac{1}{n})^n < \sum^n_{i=0} \frac{1}{i!}$, $\forall n\ge2$

I'm stuck with this proof, this is what I did: $n=2: \bigg(1+\cfrac{1}{2}\bigg)^{2} < 1 + 1 + \cfrac{1}{2}$ where verifies: $\cfrac{9}{4} < \cfrac{5}{2}$ Now with the hypothesis: $\bigg(1+\cfrac{1}{k}\bigg)^{k} < \sum_{i=0}^k \cfrac{1}{i!}$ then:…
Shiro_P
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$e(k(k−1)+1)<2^{(k−1)} \quad \forall k \geq 9$.

I am trying to solve the following problem. $e(k(k−1)+1)<2^{(k−1)} \quad \forall k \geq 9$. What have I tried so far: I tried proving this problem by induction on k. (Base Case) For k=9, we have that 198.43...<256. So Base Case holds. I am stuck on…
andimon
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