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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 $N\in\mathbb N$.

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Show by induction that $(n^2) + 1 < 2^n$ for intergers $n > 4$

So I know it's true for $n = 5$ and assumed true for some $n = k$ where $k$ is an interger greater than or equal to $5$. for $n = k + 1$ I get into a bit of a kerfuffle. I get down to $(k+1)^2 + 1 < 2^k + 2^k$ or equivalently: $(k + 1)^2 + 1 < 2^k *…
gioan
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Square of a sum equals sum of cubes

Consider a sequence $(a_n)$ of positive numbers such that $$(a_1+\cdots+a_n)^2=a_1^3+\cdots +a_n^3,\quad n\ge 1.$$ Prove that $a_n=n$ for all $n\ge 1$.
Ana María CG
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Proof $\sum\limits_{r=1}^{n} r >\frac{1}{2}n^2$ using induction

Question: $$\text{Prove by induction that, for all integers } n, n \geq 1:$$ $$\sum\limits_{r=1}^{n} r >\frac{1}{2}n^2$$ Working: Step 1 (Prove true for n=1): $$1>\frac{1}{2}(1)^2$$ Step 2 (Assume true for n=k): $$ k >\frac{1}{2}k^2$$ Step 3…
εν οίδα ότι ουδέν οίδα
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Prove by induction that $(1 + x)^n = ...$ (binomial expansion)

First, let $\binom{n}{k} = \frac{n!}{k!(n-k!)}$ for any integers $0 \le k \le n$ Show that $\binom{n-1}{k-1}+\binom{n-1}{k} = \binom{n}{k}$ for any $1 \le k \le n$ (I don't need help with this part, I have worked it out and it is true. This must…
George
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How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction?

I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $. I have managed to solve the base case, which gives 9, which is a multiple of 3. From here on, I have $(n+1)((n+1)^2 + 8)$ $n^3 + 3n^2 + 11n +…
AkshaiShah
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Basic Induction

Can someone give me a hint on how to solve this? Prove that $3^{(3n+4)} + 7^{(2n+1)}$ is divisible by 11 for all natural numbers n. So far I've got Base P(1): $3^{(3(1)+4)} + 7^{(2(1)+1)} = 2530$, which is divisible by 11 thus true P(k): $3^{(3k+4)}…
LucasCK
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Prove $a_n < 2^n$ using strong induction

Prove: $a_n < 2^n$ for all $n\ge 1$ $a_n = a_{n-1} + a_{n-2} + a_{n-3},\quad n\ge4$ $a_1=1 , a_2=2 , a_3=3$
Kayleigh
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Use the PMI to prove the following for all natural numbers $3^n≥1+2^n$

Use the PMI to prove the following for all natural numbers $3^n≥1+2^n$ Base Case: $n=1$ $3^1≥1+2^1$ $3 ≥ 3$, which is true Inductive Case: Assume $3^k ≥ 1+2^k$ [Need to Show for k+1] $3^{(k+1)} \ge 1+2^{(k+1)}$ Now from here i always get stuck…
Charlene
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Extension of induction principle proof.

i was wandering if is there an extension of the induction principle whether number the integer variables are more than 1? Example... if i need to prove that $p(n)$ is true then i would start by prove $p(0)$ and assuming for each $k <= n$ is true i'd…
user8469759
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Proving that $n \leq m^{2} \leq 2n$ by induction

I am trying to prove that there exists a perfect square between a natural number and its double, and I am trying to prove it by induction. Here is my proof so far, but I am kind of stuck. for $n=1$, choose $m=1$. For $n=2$, choose $m=2$. I figured…
ultrainstinct
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prove by induction $(1+a)^{1/n} \leq 1+a/n$ while $a\geq-1$

So, I've tried to solve this by induction, but without success. I get this equation: $(1+a)^{1/k}\leq 1+a/k$ and this equation, that I have to prove: $(1+a)^{1/(k+1)}\leq1+a/(k+1)$ I tried numerous ways, but without success, the last thing that…
user273832
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How can the proof by induction be reliable when it depends on the number of steps?

Yesterday, I got a math problem as follows. Determine with proof whether $\tan 1^\circ$ is an irrational or a rational number? My solution (method A) I solved it with the following ways. I can prove that $\tan 3^\circ$ is an irrational, the…
Display Name
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Proof by induction that $3^{n+1} > n^4$

Prove $3^{n+1} > n^4$ for all $n \in \mathbb{N}$, $n \neq 3, n \neq 4$. Let P(n) be the statement "$3^{n+1} > n^4$ such that $n \in \mathbb{N}$, $n \neq 3, n \neq 4$." I have proved the base cases, P(1), P(2), P(5), but I got stuck on the…
Tyler
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Prove $\binom{n}{k} = 0$ for $n = 0, 1, ... , k-1$

It's a homework problem. Prove $\binom{n}{k} = 0$ for $n = 0, 1, ... , k-1$ I think induction needs to be used, I can do $n = 0$ (and $n = 1$ since our teacher likes us to do the first two), but $n = m$ confuses me... Do you have to limit it so $m…
dardeshna
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Prove: $n \in Z^{\geq 2}$, $f_nf_{n+1} - f_{n-1}f_{n+2} = (-1)^{n+1}$

$n \in Z^{\geq 2}$, $f_nf_{n+1} - f_{n-1}f_{n+2} = (-1)^{n+1}$. How do you do the inductive step of this proof, every time I do it I cannot find a way to use the definition of a Fibonacci sequence to simplify the right side of the equation enough.
Katie Loss
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