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
1
vote
2 answers

Induction proof, help please?

I have a problem that I need to prove using induction. Prove that a surjective function has at least as many members in its domain as it does in its codomain. Do I begin by using the axiom of choice? Thanks.
user85362
1
vote
1 answer

Prove Recursion by Induction for Big O

I'm trying to figure out this recursive problem with induction, and I'm at a loss as to how I'm supposed to make $T(n+1) = n\log n$ , like it it wasn't $n+1$, I could do it but from what I read about induction we have to make it equal to $n\log n$.…
Mitch
  • 11
1
vote
1 answer

How to make Roger Penrose's proof of $\sqrt{2}$'s irrationality rigorous?

Taken from pg-53 of Roger Penrose's road to reality, Suppose $\sqrt{2}$ is rational, then: $$ \sqrt{2} = \frac{p}{q}$$ Where $p$ and $q$ are some integers with $ q \neq 0$: Squaring and rearrangnig: $$ 2 q^2 = p^2$$ This means $p$ is even, and…
1
vote
0 answers

induction proof of the Bernoulli inequality

I have seen different proofs of the Bernoulli inequality, but I am blocked in my solution: $${(1+x)}^n\ge 1+nx \;\forall n\ge 0$$ The base case holds, I tried when n=0; now, I have put as an induction hypothesis that the theorem holds for n-1 and to…
Lila
  • 479
1
vote
1 answer

Confusion on my inductive proof of $2^n$ ≥ $n^2$ for n ≥ 4

(The problem) Use the principle of mathematical induction to prove that $2^n$ ≥ $n^2$ for n ≥ 4 Here's my solution on paper (https://i.stack.imgur.com/iJq7M.jpg) (1) The Basis case is true: for n = 4 we have $2^4$ = $4^2$ (2) The Induction Step:…
1
vote
1 answer

Proving sum of series $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$

I am trying to prove by induction that the sum $S_n$ of the first $n$ terms of the series $$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots +\frac{1}{(2r-1)(2r+1)}+\dots $$ is $\frac{n}{2n+1}$. I have said let $n = k$.…
Steblo
  • 1,153
1
vote
2 answers

Prove $5^n > n^5 $ for $n≥6$

How can I prove $5^n > n^5 $ for $n≥ 6 $ using mathematical induction ? So my solution is Step 1: For $n = 6$ $5^6 = 15625$, $6^5 = 7776$ then $5^n > n^5 $ is true for $n = 6 $ Step 2: Assume it's true for $n = k$, $5^k > k^5 $ Step 3: Prove it's…
Seba
  • 23
1
vote
1 answer

Proof by induction - divisibility

the question goes as follows: Use proof by induction to show that $2^{n+1} + 5 \times 9^n$ is divisible by $7$. I did a few things like $f(k+1)-f(k)$ to get to $2^{k+1} + 5 \times 9^k \times 8$ but I dont understand what to do next
1
vote
1 answer

Is this mathematical induction True?

I was doing a sample paper and I came across this question which is a multiple-choice question. However, upon checking the answer after I did my calculations, one of the answers had me a little confused. Question Consider a function $f : \mathbb{N}…
Bryan Hii
  • 325
1
vote
2 answers

Help with induction proof with factorial.

I need help understanding what I am doing wrong. I need to prove (by induction): $ f(n) = \begin{cases} 1 & \text{if $n=1$} \\ (2*n - 1)*f(n - 1) & \text{if $n>=2$} \end{cases} $ $f(n) =…
Leaf313
  • 13
1
vote
2 answers

Use induction to prove that: $f^{(n)}(x) = xf^{(n−1)}(x) + (n − 1)f^{(n−2)}(x),n>2$

Let $f$ be such that $f'(x)=xf(x)$ Use induction to proove that: $f^{(n)}(x) = xf^{(n−1)}(x) + (n − 1)f^{(n−2)}(x)$ holds for all $n \geq 2$. How do I even prove that this holds for n=2? I've tried taking the second derivative and plugging in n = 2,…
1
vote
1 answer

Nested hypotheses in proof by induction

Background I have to prove that, given positive numbers $x_k$ $$ \prod_{k=1}^n x_k = 1 \Longrightarrow \sum_{k=1}^nx_k \geq n $$ where $\Longrightarrow$ means that the first condition implies the second. I already know this can be proved as a…
Andrea
  • 202
  • 1
  • 10
1
vote
1 answer

Proving by induction $\sum_{j = b}^a\frac{j!}{b!(j-b)!} = \frac{(a+1)!}{(b+1)!(a-b)!}$

I am only recently studying how to write proofs, and as an exercise I set off to prove the following equation: \begin{equation} \sum_{j = b}^a\frac{j!}{b!(j-b)!} = \frac{(a+1)!}{(b+1)!(a-b)!} \end{equation} The strategy I chose for proving this is…
Werther
  • 85
1
vote
0 answers

Validity of a proof of Weak to Strong Induction

http://mathcenter.oxford.emory.edu/site/math125/strongInductionEquivalence/ In the link given above, I'm not sure that the proof of 'whenever weak induction holds, strong induction holds' is correct. This is because in the proof the second…