This article on the demonstration of paradoxes and fallacies in mathematics was written by Matrix67, Chinese Department of Peking University. It feels interesting to read, so I will share it with you and have a brainstorm.

1=2? Classic "proof" in history

Let a = b, then AB = A 2, subtract B 2 from both sides of the equal sign to get AB-B 2 = A 2-B 2. Note that the left side of this equation can be given a b, and the right side is the square difference, so there is b (a-b) = (a+b) (a-b). If (a-b) is omitted, there is b = a+b However, a = b, so b = b+b, that is, b = 2b. If b is omitted, 1 = 2 is obtained.

This is probably the most classic fallacy ever. Jiang Fengnan wrote in his short science fiction novel Divide by Zero:

cite

There is a well-known "proof" that one equals two. It begins with some definitions: "Let a =1; Let b = 1. " It ends with the conclusion "a = 2a", that is, one equals two. Hidden in the middle and inconspicuous is a division divided by zero. At this point, the proof has left the edge and made all the rules invalid. Allowing division by zero can not only prove that one and two are equal, but also prove that any two numbers-real or imaginary, rational or irrational-are equal.

The problem of this proof must be clear to everyone: both sides of the equal sign cannot be divisible by a-b at the same time, because we assume that a = b, that is, a-b equals 0.

Power of infinite series (1)

When I was in primary school, this question puzzled me for a long time: What is the following formula?

1 + (- 1) + 1 + (- 1) + 1 + (- 1) + …

On the one hand:

1 + (- 1) + 1 + (- 1) + 1 + (- 1) + …

= [ 1 + (- 1)] + [ 1 + (- 1)] + [ 1 + (- 1)] + …

= 0 + 0 + 0 + …

= 0

On the other hand:

1 + (- 1) + 1 + (- 1) + 1 + (- 1) + …

= 1 + [(- 1) + 1] + [(- 1) + 1] + [(- 1) + …

= 1 + 0 + 0 + 0 + …

= 1

Doesn't this mean 0 = 1

Later, I learned that this formula can also be equal to 1/2. Let s =1+(-1)+1+(-1)+…, then S = 1-S, and the solution is S = 1/2.

After studying calculus, I finally understand that this infinite series is divergent and there is no so-called "sum". What is the result of adding infinite numbers? This needs to be defined

The Power of Infinite Series (2)

The same trick can produce more incredible things. For example, make

x = 1 + 2 + 4 + 8 + 16 + …

There are:

2x = 2 + 4 + 8 + 16 + …

So:

2x-x = x =(2+4+8+ 16+……)-( 1+2+4+8+ 16+……)=- 1

That is to say:

1 + 2 + 4 + 8 + 16 + … = - 1

Square root diagram (1)

Theorem: All numbers are equal.

Proof: Take any two numbers A and B and let t = A+B. Therefore,

a + b = t

(a + b)(a - b) = t(a - b)

a^2 b^2 = t a-t b

A^2-Toronto = b^2-Toronto

a^2 - t a + (t^2)/4 = b^2 - t b + (t^2)/4

(A-T/2) 2 = (B-T/2) 2

a - t/2 = b - t/2

a = b

What's going on here?

The problem lies in the penultimate line.

Always remember that x^2 = y^2 does not mean x = y, only x = y.

The plot of the square root (2)

1 = √ 1 = √(- 1)(- 1) = √- 1 √- 1 = - 1

Hmm?

Only when both x and y are positive numbers, √√ x y = √√√ x y is valid.

There are two square roots-1, i and -i. √(- 1)(- 1). After expansion, write I (-I), which is exactly equal to 1.

Plural number is king.

Consider this equation

x^2 + x + 1 = 0

The transferred projects are as follows

x^2 = - x - 1

When both sides of the equation are divided by x at the same time, there are

x = - 1 - 1/x

Substituting the above formula into the original formula, there are

x^2+(- 1- 1/x)+ 1 = 0

that is

x^2 - 1/x = 0

that is

x^3 = 1

That is, x = 1.

X = 1 returns to the original formula and gets12+1+1= 0. In other words, 3 = 0, hehe!

Actually, x = 1 is not the solution of the equation x 2+x+1= 0. In the real number range, the equation x 2+x+1= 0 has no solution, but there are two solutions in the complex number range.

On the other hand, x = 1 is just one of the solutions of x 3 = 1. X 3 = 1 Actually, a * * * has three solutions, but the other two solutions are all in the range of complex numbers. Considering that the equation x 3-1= (x-1) (x 2+x+1) = 0, it is easy to see that the two complex solutions of x 3 =1are exactly x 2+x+1. Therefore, it is not contradictory that x 2+x+1= 0 and X 3 = 1 hold simultaneously.

Note that once the plural is introduced, this fallacy has a complete and beautiful explanation. Perhaps this also shows the necessity of introducing the concept of complex number.

A funny mistake.

As we all know,

1 + 2 + 3 + … + n = n(n+ 1) / 2

We use n- 1 instead of n.

1+2+3+…+(n- 1)=(n- 1)n/2

Add 1 to both sides of the equation and you will get:

1+2+3+…+n =(n- 1)n/2+ 1

namely

n(n+ 1)/2 =(n- 1)n/2+ 1

Have after expansion

n^2 / 2 + n / 2 = n^2 / 2 - n / 2 + 1

It can be seen that n = 1 is the only solution of this equation.

That is to say,1+2+3+…+n = n (n+1)/2 is only valid when n = 1

There is a very hidden and funny mistake in this reasoning process. After adding 1 to both sides of the equation, the left side of the equation should be

1+2+3+…+(n-2)+(n- 1)+ 1

1 yuan equals 1 min?

I want to empty your wallet with the power of mathematics! Please see:

1 yuan = 100 point = (10 point) 2 = (0. 1 yuan) 2 = 0.0 1 yuan = 1 point.

I have been trying to deceive children with this, because primary school (or even middle school) education ignores a very important idea: units should also participate in the operation. In fact, "100 = (10) 2" is not valid, and the square of "100" should be "100", just as the square of "100" is "/kloc-0".

The Tragedy of Mathematical Induction (1)

The following "proof" was given by the mathematician George Boya: Given any N horses, it can be proved that all the N horses are of the same color.

Induction n: First of all, when n = 1, the proposition is obviously established. If the proposition holds for n = k, consider the case of n = k+ 1: because {# 1, # 2, ..., # k} these k horses are the same color, {# 2, # 3, ..., # k+ 1} these k horses.

The mistake of this proof is that n = 2 cannot be deduced from n = 1, although this induction is correct when n is large. This is a strange example of a mistake in mathematical induction: there is nothing wrong with the basic situation and inductive reasoning, but it is stuck at a certain step in the inductive process.

The Tragedy of Mathematical Induction (2)

Next, let me prove to you that all positive integers are equal.

In order to prove this, it is only necessary to explain that for any two positive integers a and b, there is a = b.

To prove this, it is only necessary to explain that for all positive integers n, if max(a, b) = n, then a = b.

We generalize n. When n = 1, because both A and B are positive integers, both A and B must be equal to 1, so A = B. If n = k, the proposition holds. We assume that max(a, b) = k+ 1. Then max(a- 1, b- 1) = k, and from the inductive hypothesis, a- 1 = b- 1, that is, a = b.

The problem is that a- 1 or b- 1 may not be a positive integer, so the inductive hypothesis cannot be applied.