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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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Prove that the identity is true for all natural numbers

for the identity: $$\frac{n!}{x(x+1)(x+2)...(x+n)} = \frac{A_0}{x+0} + \frac{A_1}{x+1}+...+ \frac{A_n}{x+n}$$ prove $$A_k= (-1)^kC(n,k)$$ I think this might work by induction, but i am not able to arrive at a final answer. please help!!!
Coherent Sheaf
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Induction and contradiction

I want to prove a statement by induction , I tasted the base case, then I considered the induction hypothesis for $n$ , so I assumed by absurdity that is not true for $n + 1$ but this contradicts the induction hypothesis , this proof is correct ?…
Israel Meireles Chrisostomo
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How was induction used here?

I was reading the proof of the induction for this question and didn't understand the step where they did $t! > (n^2)^{t-n^2} = n^tn^{t-n^2} > n^t$. How is $n^2$ even related to the problem and if so how is the inequality even true? Problem and…
user19405892
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induction (vs recursion) in proof

This post is about "proof by induction". I want to understand if I've been doing it right. I'm a little confused by something I read in a textbook. My background in math doesn't go beyond undergraduate classes (basic analysis, algebra). On pg47 of…
confusedagain
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Use induction to prove that $\sum_{k=1}^{2^{n}}\frac{1}{k} \geq 1 +\frac{n}{2}$?

For the base case I have $n = 1$, so ${1\over1} + {1\over2} = {3\over2}, 1 + {1\over2} = {3\over2}$. Hence this is true. For the induction hypothesis we can assume n=k is true also. For the induction step I plug in $n=k+1$ but i'm not sure what to…
bella
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Show that the sequence $(x_n)=c^{\frac{1}{n}}$ is increasing for $0 < c < 1$.

Show that the sequence $(x_n)=c^{\frac{1}{n}}$ is increasing for $0 < c < 1$. I am trying to do this using induction. We see for the base case that $x_1 = c$ and $x_2 = c^{0.5}$, so clearly $x_1 < x_2$. For the induction step, we assume $x_n <…
flubsy
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Prove by induction: $\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$

$\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ For $n=1$ equality is true. For $n=m$ $m-{m\choose 2}\frac{1}{2}+...+(-1)^{m+1}\frac{1}{m}=1+\frac{1}{2}+...+\frac{1}{m}$ For…
user300045
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Prove that $2^n>n^4$ for all $n\geq 17$

I'm always a bit fuzzy on how to solve induction problems involving inequalities. I've managed to get somewhere though, but it looks like I have to go down four levels of induction to prove. This is what I have so far: Base…
Mirrana
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Proof by induction on n

So I aim to prove that $n^2 \leq 2^n + 1$ for all integers $n \geq 1$. We can see that this is true for $n=1$ since $1 \leq 3$. Now I suppose that this is true for an arbitrary $k$ such that $k \geq 1$. So $k^2 \leq 2^k +1$. From this, I want to…
user247618
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Bernoullis inequality proof

Hi I'm asked to do a proof for bernoullis inequality which is $(1+a)^n \geq 1+na$ where $a\geq-1$ I'm proving by induction by the way. So far these are my steps $(1+a)^1 \geq 1+a$ Then $(1+a)^k \geq 1+ka$ This is where I get lost. I thought I was…
user3258845
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Proving $\frac{n^n}{3^n} < n!$ for $n\ge6$ by induction

How would I prove this using mathematical induction: $\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$. Here is what I have tried: $\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$ Base case: $\dfrac{6^6}{3^6} < 6!$. Yes, this is true. Assume for $k$ it's true so…
CodeR
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$9 \mid 4n^2 + 15n - 1$ for $n \in \mathbb N$

How to prove by induction that $9 \mid 4n^2 + 15n - 1$ for every $n \in \mathbb N$? For $n = 1$ $4 \cdot 1^2 + 15 \cdot 1 - 1 = 18$ For $n \ge 2$ If $4n^2 + 15n - 1 = 9k$ then $4(n+1)^2 + 15(n+1) - 1 = 4n^2 + 23n + 18 = 9k + 8n + 19$
user4201961
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Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$

I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understand? $$\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} =…
MathsNoob2015
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Basic mathematical induction regarding inequalities

These are just the examples from my textbook, but I don't think it did not explain well. One of the problem was to prove the inequality $$n<2^n$$ for all integers $n$. I understand we assume if $P(n)$ is true, $P(k+1)$ is true. Thus $k+1<…
Minjae
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