Which two irrational numbers add up to five?

Which two irrational numbers add up to five?

For example, if a is the square of 2, b is 5-a, then a+b=5, and a, b are irrational numbers.
Let a be an arbitrary irrational number, b be a 5-a, obviously a+b=5. Here we can prove that b is also an irrational number, because if b is not an irrational number, then from the closure of the addition and subtraction operation of rational number, we can know that 5-b is a rational number, i.e., a is a rational number, which contradicts the definition of a, so b is also an irrational number.

How could they prove that the sect was unreasonable and surpassed? At the same time, he could explain how to surpass numbers. On the 1st and 2nd floors, there was no level at all. The fourth floor was so strong, but he hoped to use the brackets to cover it clearly, so he didn't quite understand How do we prove that pie is irrational and transcendental? By the way, explain the transcendental number. Don't make a long speech. There's no level on the 1st and 2nd floors. The fourth floor is strong, but I hope it's clear in brackets. I do n' t understand How do you prove that pie is irrational and transcendent? By the way, explain the transcendental number. Don't give me an irrelevant long speech. There's no level on the 1st and 2nd floors. The fourth floor is strong, but I hope it's clear in brackets. I do n' t understand

The individual formula was changed more clearly.
A transcendental number is the part of a real number that can not be expressed as the root of an algebraic equation.
In real numbers, algebraic numbers are countable, so transcendental numbers are uncountable.
It's much easier to prove Pi is an irrational number, and I' ve forgotten what I've seen before. Here's a proof from the Internet:
This proof belongs to Ivan Niven. assumption pi=a/b, which we define (for a certain n):
F (x)=(x^n)*(a-bx)^n /n!
F (x)= f (x)+...+(-1)^j*f^(2j)(x)+...+(-1)^n*f^(2n)(x)
Where f^(2j) is the second derivative of f.
Then f and F have the following properties (both easily verified):
1) F (x) is an integral coefficient polynomial divided by n! .
2) F (x)= f (Pi - x)
3) F increases strictly in (0, pi) interval, and f (x) tends to 0 when x tends to 0,
When x tends to pi, f (x) tends to pi^n*a^n/n!
4) For 0= n, the jth derivative of f is an integer at 0 and pi (see 1).
6) F (0) and F (pi) are integers (4),5).
7) F + F "= f
8)(F'·sin-F·cos)'=f·sin (see 7).
Thus, a definite integral of f·sin from 0 to pi is
(F'(pi) sin (pi)-F (pi) cos (pi))-(F'(0) sin (0)-F (0) cos (0))
=F (pi)+F (0)
This is an integer.
The problem is that if n is obtained very large, the definite integral of f·sin from 0 to pi must be strictly greater than 0 and strictly less than 1.
Lindemann first gave the proof that Pi is a transcendental number. I haven't studied algebraic number theory. I have to use The Hermite - Lindemann Transcedence Theorem. may have in books on algebraic number theory. Some books on modern geometry also have the proof of Pi's transcendence in the question of turning a circle into a square.
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