Reduction to absurdity

How can I direct proof?

  • Reduction to absurdity:proof by contradictionThe need toAssume that the statement to be proved is false, and then derive a contradiction so that the proof cannot be false
  • , etc.

Example of reduction: Prove that 2\sqrt22 is an irrational number.

Solution: The direct proof is difficult, because the proof of irrational numbers is by proving that there are no integers a and b, such that the number =a/b. 2=ab\ SQRT {2}= cfrac{a}{b}2 =ba, a2=2b2a^2=2b^2a2=2b2 So both a2a ^ 2 a2 is a multiple of 2, so is a multiple of 2, namely a can be divided exactly by 2, writing ∣ a2 | 2 a2 ∣ a. If a is even, So both a2a ^ 2 a2 is a multiple of 4, namely 4 ∣ a24 | a ^ 24 ∣ a2, namely 4 ∣ 2 b24 24 ∣ | 2 b ^ 2 b2, namely 2 ∣ b22 | b ^ 22 ∣ b2, so the same, b is an even number, then got the contradiction – ab \ cfrac {a} {b} ba isn’t about fractions. So 2\sqrt22 is irrational.

Mathematical induction

Induction axiom: Make P(n)P(n)P(n) predicate. If P (0) P (0) P (0) is true, and for all natural Numbers n, P (n) – > P (n + 1) establishment of P (n) – > P (n + 1) establishment of P (n) – > P (n + 1), so for all natural Numbers, P (n), P (n), P (n) is established.

A predicate is, in the first place, a proposition, and, in the second place, the correctness of the proposition depends on some variable

Why does the axiom of induction hold?

Because if P(0) is true,P(0)->P(1), then P(1) is true; P(1)->P(2), so P(2) is true… ; And so on…

To prove by induction:
For all n > 0 . 1 + 2 + 3 + . . . + n = n ( n + 1 ) 2 For all n>0, 1+2+3+… +n=\cfrac{n(n+1)}{2}

  1. Basecase: If n=1,sum(n)=1
  2. 1. The human revolution step If k, the sum (k) = k (k + 1) 2 was established, then the k + 1, sum (k + 1) = sum (k) + k + 1 = (k + 1) (k + 2) 2 sum (k) = \ cfrac {k (k + 1)} {2} was established, then the k + 1, The sum (k + 1) = sum (k) + k + 1 = \ cfrac {(k + 1) (k + 2)} {2} sum (k) = 2 k (k + 1) was established, then the k + 1, sum (k + 1) = sum (k) + k + 1 = 2 (k + 1) (k + 2), meet the same type
  3. So the

(Using the above axiom of induction, where P(n) is 1+2+3+… +n=n(n+1)21+2+3+… +n=\cfrac{n(n+1)}{2}1+2+3+… +n=2n(n+1)

To prove by induction:
For all natural numbers n . ( n 3 – n ) is 3 A multiple of . namely 3 ∣ ( n 3 – n ) For all natural Numbers n (n ^ 3 – n) is a multiple of 3, namely 3 | (n ^ 3 – n)

The process is slightly simple

Prove by induction that all horses are the same color that all horses are the same color that all horses are the same color that all horses are the same color (of course, this proof is wrong and we need to find out what is wrong)

The statement that all horses have the same color, we need to find a way to translate it into this predicate, which is to find a variable that makes this statement true in relation to that variable, and that variable tends to infinity which will ultimately give us the proof of our problem.

It is easy to think of an appropriate predicate P(n): in any set of n horses (n>=1), all the horses in the set are of the same color

  1. Prove BaseCase basic case:

Any collection with only one horse, of course, only one color

  1. 1. The human revolution step

If P(n) is true, we need to prove that P(n+1) is true.

Prove that P(n+1) is true, that is, prove that H(1),H(2)… H of n plus 1 has the same color. We know that any n horses have the same color, so H(1),H(2)… H(n) has the same color, and H(2),H(3)… H(n+1) has the same color, so H(1),H(2)… H(n+1) has the same color, that is, P(n+1) is true

What’s the problem?

In the Inductive step of the revolution,H(1~ N) is the same color, and H(2 ~ N +1) is the same color, so H(1~ N +1) is the same color. This is true if (1 ~ n) and (2 ~ n+1) have an intersection, but when n=1, there is no intersection, so there is an error when P(1) pushes P(2), and P(1) cannot deduce P(2).

Finally: prove by induction: there is a side of length 2n and there is a side of length 2^n and there is a side of length 2n