Mathematics Stack Exchange
Mathematics Stack Exchange

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
0
votes
1 answer

Induction Proof with Binomial Coefficient

For all n > 0 prove that ${2n}\choose{n}$ < $4^n$ It seems like a simple proof but I'm not sure how to continue it. What I have so far: Base Case: Let n=1 ${2}\choose{1}$ = 2 $4^1$ = 4 2 < 4 Inductive Step: Assume ${2(x+1)}\choose{x+1}$ <…
JanoyCresva
  • 486
  • 7
  • 18
0
votes
1 answer

Induction with Fibonacci Identity

Using induction prove that: $$\sum_{i=0}^{2n} (-1)^iF_i = (F_{2n-1})-1$$ I am on the inductive step but am stuck with how to proceed. This is what I have after the base case so far: $$\sum_{i=0}^{2n+2} (-1)^iF_i = (F_{2n+2})-1$$ …
JanoyCresva
  • 486
  • 7
  • 18
0
votes
2 answers

math induction of sin(x)-sin(3x)...

how would you use induction to prove this: $\sin(x)-sin(3x)+sin(5x)-...+(-1)^{(n+1)}sin[(2n-1)x] = \frac{(-1)^{(n+1)}sin2nx}{2cosx} $ I know how you assume its true for n=k, and then prove for n=k+1, but I get to Left Hand Side:…
Smithy
  • 51
0
votes
5 answers

Prove by induction that $3^{n^{2}}+9n+6$ is divisible by 6.

I am having trouble with the induction part of the proof that the expression bellow is divisible by 6 $$ 3^{n^{2}}+9n+6 $$ $n->n+1$ $$ 3^{n^{2}+2n+1}+9n+9+6 $$ I have been able to see that this is divisible by $3$, so if I would be able to find…
Jure Vreča
  • 139
0
votes
1 answer

Performing Induction on the process of Induction (Function Spaces?)

I think it was Gauss who taught us that $$1+2+...+n = \frac{n(n+1)}2$$ for natural numbers. We can easily verify this with a proof by induction. However, what if I would like to find the sum of all the natural numbers between any two given natural…
DaFreed
  • 1
  • 1
0
votes
1 answer

Stuck in inductive step

Statement Use mathematical induction to show that given a set of n + 1 positive integers, none exceeding 2n, there is at least one integer in this set that divides another integer in the set. It's clear that the set can't contain two equal integers…
Antonio González Borrego
  • 89
0
votes
4 answers

What's wrong with this induction proof?

Theorem: $n < 10^n$ For $P(0)$, $0 < 1$ Inductive Hypothesis: Assume $k < 10^k$ We must show $k+1 < 10^{k+1} $ By the inductive hypothesis, we know, $k < 10^k$ Plugging in $k + 1$ in place of $k$ for the inductive hypothesis, we get $k+1 < 10^{k+1}…
cprogrammer
  • 153
  • 2
  • 7
0
votes
1 answer

Induction for $1^2+3^2+5^2+...+(2n-1)^2= \binom{2n+1}{3}$

Induction for $1^2+3^2+5^2+...+(2n-1)^2= \binom{2n+1}{3}$. I am stuck... This is what I have so far... Base case: $n=1, 1=\binom{2n+1}{3}\rightarrow 1=1$. Inductive step: Assume $n=k$ is true and show $n=k+1$…
MathIsHard
  • 2,733
  • 16
  • 47
0
votes
2 answers

If $P_n$ is the statement $3^n>n,$ prove $P_{n+1}$ is true whenever $P_n$ is true.

If $3^n>n$ then $3^{n+1}>n+1$ I could not solve this problem
bappy
  • 91
0
votes
1 answer

Prove matrix identity by induction

Let $\lambda_1=\frac{1-\sqrt{5}}{2}$ og $\lambda_2=\frac{1+\sqrt{5}}{2}$. Show by induction in $k$ that \begin{equation*} A^k=\frac{1}{\sqrt{5}} \begin{pmatrix} \lambda_2^{k-1}-\lambda_1^{k-1} & \lambda_2^k-\lambda_1^k\\ …
user388910
0
votes
2 answers

Prove by induction question

Please help me to prove this:$$\sum_{r=1}^n \frac{1}{\sqrt r} > \sqrt n$$ for $$n \ge 2, n \in \Bbb Z$$ I finished the first step with this: $$\frac {2+\sqrt 2}{2}> \sqrt 2 \rightarrow True$$ But I am not sure whether this is correct. Then I don't…
SarfIns
  • 25
0
votes
2 answers

Prove $a^n

Can I prove $a^n
user401653
  • 187
0
votes
1 answer

How Can I prove that $x_1^2+x_2^2+......+x_n^2>0$ by matematical induction

The problem is this: Prove by induction that $x_1^2+x_2^2+.....+x_n^2>0$ unless $x_1=...=x_n=0. $ I have tried: If $x_n$> 0 then For $n=1$: $$x_1^2>0$$ We suppose that it is true for $n=k$:$$ x_1^2+x_2^2+...+x_k^2>0$$ We will prove for…
Alexis Galois
  • 187
0
votes
9 answers

Is $n^{2} + 5n + 1$ ever an even integer for a chosen natural number n?

How is it shown that this number is odd for all natural numbers? Can someone show me why this is not an even integer using induction? Thank you in advance.
Garrett
  • 397
0
votes
4 answers

Proof by mathematical induction that expression is divisible by expression,

The question is : Prove by mathematical induction that : $$ a^{2n-1} + b^{2n+1} $$ is divisible by $$a+b$$ I've done a lot of stuff but can't put them down in tex properly. Thanks.
Mob
  • 375
1 2 3
…
99 100