Does anyone know any simple examples to explain induction. Ideally some sort of reasoning that they would come across in normal life?
I'm convinced that induction is extremely obvious and in the right context it'd be obvious to anyone normal person but I can't find a good example, non stupid.
Induction is just proving that if something is true in one case, this implies that it is true in the next case. Example:
Prove the sum from 1 to n is n(n+1)/2
1. Prove this is true in "one case" aka the base case: the sum from 1 to 1 is 1 = 1(2)/2
2. Prove that if the sum from 1 to n is n(n+1)/2 then the same must be true for n+1:
if the sum from 1 to n is n(n+1)/2 for some n then the sum from 1 to n+1 is n(n+1)/2+n+1
n(n+1)/2+n+1 = n(n+1)/2+2(n+1)/2 = (n+2)(n+1)/2 = (n+1)(n+1+1)/2
So the sum from 1 to n+1 is (n+1)(n+1+1)/2 if the sum from 1 to n is n(n+1)/2 for some n
The sum from 1 to n is n(n+1)/2 for n=1, so it is true for 1+1 = 2
The sum from 1 to n is n(n+1)/2 for n=2, so it is true for 2+1 = 3
ad infinitum
Thus the sum from 1 to n is n(n+1)/2 for all positive integers.
That's a really nice gif, do you mind if I save it?
Simplifyeth not induction thee chop-logic. Its a straight-f'rward concept, if 't be true ingraft'rs receiveth not t t's because those gents art uncultur'd swine
If I have to explain induction to plebs I'm not going to use the sum of all natural numbers? The numbers must be hidden under the details.
Suit yourself
If you go to a grocery store to buy bread and notice that the tags are colour coded by expiration date do you assume that some date will always correspond to that colour? Induction isn't really a real world thing because people are people but this is something I induct upon to quicken my bread search when I shop. There's no real reason to change the colour coding so induction isn't unreasonable
There is a book called "A Journey into Mathematics" by Joseph Rotman. It has one of the best explanations of induction I've seen, with simple and complicated examples.
Wanna give the gist of the simple ones?
Just read the first chapter.
If you can't do that, Eric Carle wrote a very interesting text on algebraic topology and number theory you might be interested in. I can't remember the title though.
So:
If the next step above a white step is white, and the first step is white, then all the steps are white.
Or if you found a trail of candy, but knew some of them were poisoned, then there'll be a first poisoned candy.
Isn't it just concatenated modus ponens?
The rule
(F, F⇒G)/G, where ⇒ means "implies, " which is the sole rule of inference in propositional calculus. This rule states that if each of F and F⇒G is either an axiom or a theorem formally deduced from axioms by application of inference rules, then G is also a formal theorem.
1/2 + it
This post reaks of autism
I don't understand what's so fucking hard about induction. Its literally if P(k+1) when P(k) is true, then P(n) is true for all natural numbers n.
[math]P(0)+n P'(0)+\frac{1}{2} n^2 P''(0)+\frac{1}{6} n^3 P^{(3)}(0)+\frac{1}{24} n^4 P^{(4)}(0)+O\left(n^5\right)[/math]
Here's something simple that I had to use earlier.
[eqn]10^n\cong1(3), \forall n \in \mathbb{Z},n\geq 0 [/eqn]
Basically this Some other good, simple examples are one that was on some homework for introductory analytical calculus and the first midterm.
>A) Use the principle of mathematical induction to prove that (12^n) - (5^n) is a multiple of 7 for all n within the set of natural numbers.
f(n) = (12^n) - (5^n) =7k
f(1) = (12^1) - (5^1) = 12-5 = 7(1)
f(n+1) = (12^(n+1)) - (5^(n+1))
=> (12^(n+1)) -12(5^n) +12(5^n) - 5(5^n)
=> 12((12^n) -(5^n)) +7(5^n)
=> 12(7k) +7(5^n)
=> 7(12k) +7(5^n)
=> 7(12k +t^n)
=> 7(k*)
>B) Use the principle of mathematical induction to prove that f(n) = (2^0) +(2^1) +(2^2) +... +(2^(n-1)) = (2^n) -1 is true for all n within the set of natural numbers
f(n) = (2^0) +(2^1) +(2^2) +... +(2^(n-1)) = (2^n) -1
f(1) = (2^(1-1)) = (2^0) = 1 = 2 - 1 = (2^1) -1
f(n+1) = (2^0) +(2^1) +(2^2) +... +(2^n) = (2^(n+1)) -1
=> (2^0) +(2^1) +(2^2) +... +(2^(n-1) +(2^n) = (2^n) -1 +(2^n)
=> (2^0) +(2^1) +(2^2) +... +(2^(n-1) +(2^n) = 2(2^n) -1
=> (2^0) +(2^1) +(2^2) +... +(2^(n-1) +(2^n) = (2^(n+1)) -1
1x3?
1mod3 thought it was clear with the congruence sign
That looks more like "approximately equal" to me! The same symbol can have lots of uses.
Eric Carle?
x * y = x + y
x*y = x|0, y|0
x+y= x|2, y|2
There is a difference.
assume that there's a line of standing domino stones, perhaps neverending. if you know two things, namely
> 1) someone will tip the first domino stone over
> 2) each domino stone will tip over the one domino stone directly after it when it falls down
then induction is nothing but saying "well, then each and every domino stone will fall over eventually!"
[math]\approx[/math] is the approximately equal sign
hmm, at our uni we use that sign to represent isomorphic, and 3 parallel lines to represent congruence. Is there no international convention for these things??
Pretty much.
If a black guy only fucks black guys. And you have a line of gays. If the first guy is black, then so is the second, the third, .... By induction all of the guys in the line are black.
Is unity having an episode?
let's say you live in a town where everybody is really self conscious and insecure - if they see someone wearing a hat, they'll immediately wear one too, and vise versa.
now let's say that you're a shopkeeper on his way to the shop and your wife calls your from the shop and tells you there's a massive line of people at the entrance (it's a straight line, everyone can only see the person in front of them). For some reason you want to know whether the people in line are wearing hats or not (maybe you run a hat shop), what should you ask your wife?
just ask her if the first person in line is wearing a hat - if he is, then the person behind him (that sees him) will automatically wear one too, and from the same reason the one after him as well etc.
that's basically induction - you prove that if the case holds for a certain member of a well ordered group (let's say n), it must hold for the next one in order (n+1). then, by checking the first member (the "smallest" in order), you hereby prove that the statement holds for the entire group.
hope this helps.
3 straight lines is generally what i use but i couldn't find a symbol for it latex at the time, I actually just found it it's \equiv
She is the episode.