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

Proof by induction that if $\text{five}(0) = 10$ and $\text{five}(n+1) = \text{five}(n) + 5$ then $\text{five}(j) = 5(j+2)$ for all $j$

Let the function ${\rm five}(n)$ be a function defined by the two equations: ${\rm five}(0) = 10 \\ {\rm five}(n+1) = {\rm five}(n) + 5$ Prove that: ${\rm five}(j) = 5*(j + 2)$ for all non-negative integers $j$. I asked a friend today a question…
user119325
  • 127
  • 1
  • 6
0
votes
1 answer

proof by induction, how to start with the specific example. 5^n - 1 is divisible by 4

I am looking at proof by induction as part of my maths module for my upcoming examination. I ave worked through several problems of induction, but i am not yet fully capable. The problem i have been asked is: for every n greater of equal to 1 show…
0
votes
2 answers

Is my mathematical induction answer ($n < 3^n$) correct or not?

Using Mathematical induction, prove that, for each $n \in \aleph$, $n < 3^n$. I prove that using following steps. Please tell me my answer is correct or not? $$n < 3^n$$ for $n = 0$, $n < 3^n \iff 0 < 3^0$ for $n = k$, $k < 3^k \iff 0 < 3^0$ for…
0
votes
1 answer

Induction over reals

Would this argument be valid for proving A for all real numbers greater than or equal to a: Prove that A is true at n = a. Assume that A is true for all $ a\le n < k$. Prove that A is true at k.
0
votes
2 answers

Induction Problem: $\sum_{j=1}^{n}(-1)^j\binom{n}{j}\frac{1}{j+1}=-\frac{n}{n+1}$

Prove that: $$\sum_{j=1}^{n}(-1)^j\binom{n}{j}\frac{1}{j+1}=-\frac{n}{n+1} $$ I know that 1.I need prove it by induction 2.this can be helpful : $ n\binom{n-1}{j-1}=j\binom{n}{j} $ whatever, i tried to do it by induction and i don´t know how i can…
Luis GC
  • 412
0
votes
4 answers

Prove by using Mathematical induction (sum of the first $n$ odd numbers is $n^2$)

$$1+3+...+(2n-1) = n^2 for\quad all\quad n∈N$$ Been watching youtube vidoes but still confused. Step 1: Show that n=1 is true (Initial value) LHS = 2(1)-(1) = 1, RHS = $1^2$=1 therefore LHS=RHS. N=1 is true. Step 2: Assume n=k is…
0
votes
1 answer

Prove by mathematical induction for all n in set N

Consider the sequence $\{a_n\}$, where $a_0 = 1$; $a_1 = 2$; $a_2 = 3$ and $a_n = a_{n-1} + a_{n-2} + a_{n-3}$; $n$ belongs to set $\mathbb{Z}^+$; where $n \ge 3$. (b) Prove by mathematical induction that for all $n \in \mathbb{N}$ , we have…
0
votes
1 answer

proof for the alternate version of strong induction

Prove: Let $t$ be a fixed integer and $j$ a fixed positive integer. Show that if $P(t), P(t+1),\dotsc,P(t+j)$ are true and $[P(t)\land P(t+ 1) \land \dotsb \land P(k)]\implies P(k+1)$ is true for every integer $k \ge t$, then $P(n)$ is true for…
0
votes
2 answers

Using induction to prove $a_n >2^n$

For the sequence $a_n=2a_{n-1}+1$ where $a_0=1$ Show that $a_n>2^n$ using induction. Use proof by contradiction (minimum counterexample). Attempt: 1. I assume, that $a_n>2^n$ as my induction hypothesys. Now, I try to show that…
Koba
  • 1,233
0
votes
1 answer

induction triangle question prove x triangles are formed

Prove with induction: Given n non parallel lines such that no three intersect at a point, there are c(n,3) triangles formed. so I have n!/(n-3)!3! triangles if this condition holds, that is all I understand how to do from this question
SSS
  • 41
0
votes
1 answer

Show that $2^{2^{n+1}-1} +1 \le 2^{2^n}$

I tried an inductive proof but it didn't work out and I'm stuck for ideas.
0
votes
2 answers

Prove by induction that a^n+a^-n is an integer.

I am to prove by induction that given that $a+1/a$ is an integer (i.e. belongs to Z ) then $a^n+1/a^n$ is an integer too. I'm pretty much clueless here. Thanks in advance.
Danny
  • 189
  • 2
  • 10
0
votes
1 answer

Prove by induction on strings

I have this question: Prove by induction on strings that for any binary string $w$, $(oc(w))^R = oc(w^R)$. note: if $w$ is a string in $\{1,0\}^*$, the one's complement of $w$, $oc(w)$ is the unique string, of the same length as $w$, that has a…
john
  • 11
0
votes
1 answer

Need Help Solving Polynomial Equation

I'm working on an induction problem that basically boils down to this equation: $$2(-1)^k+ 6(2^k)\left(-\frac{1}{2}\right)^{k+1} + (-1)^{k}=0$$ I'm fairly confident that the equation above is the solution to the problem, but I am unable to simplify…
0
votes
1 answer

Inductive proof for all naturals (including 0), x, starting from one of two given starting points

The given starting points for this inductive proof are the following: 1) the formulas, G(2x-2) = G(x)^2 - G(x-2)^2 and, G(2x-1) = G(x+1)G(x) - G(x-1)G(x-2) or 2) the single formula for G(x) given above, in terms of the number a = ((5^.5) -…