lean2/library/theories/number_theory/irrational_roots.lean
2015-12-05 23:50:01 -08:00

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/-
Copyright (c) 2015 Jeremy Avigad. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Author: Jeremy Avigad
A proof that if n > 1 and a > 0, then the nth root of a is irrational, unless a is a perfect nth power.
-/
import data.rat .prime_factorization
open eq.ops
/- First, a textbook proof that sqrt 2 is irrational. -/
section
open nat
theorem sqrt_two_irrational {a b : } (co : coprime a b) : a^2 ≠ 2 * b^2 :=
assume H : a^2 = 2 * b^2,
have even (a^2),
from even_of_exists (exists.intro _ H),
have even a,
from even_of_even_pow this,
obtain (c : nat) (aeq : a = 2 * c),
from exists_of_even this,
have 2 * (2 * c^2) = 2 * b^2,
by rewrite [-H, aeq, *pow_two, mul.assoc, mul.left_comm c],
have 2 * c^2 = b^2,
from eq_of_mul_eq_mul_left dec_trivial this,
have even (b^2),
from even_of_exists (exists.intro _ (eq.symm this)),
have even b,
from even_of_even_pow this,
assert 2 gcd a b,
from dvd_gcd (dvd_of_even `even a`) (dvd_of_even `even b`),
have 2 1,
begin rewrite [gcd_eq_one_of_coprime co at this], exact this end,
show false, from absurd `2 1` dec_trivial
end
/-
Replacing 2 by an arbitrary prime and the power 2 by any n ≥ 1 yields the stronger result
that the nth root of an integer is irrational, unless the integer is already a perfect nth
power.
-/
section
open nat decidable
theorem root_irrational {a b c n : } (npos : n > 0) (apos : a > 0) (co : coprime a b)
(H : a^n = c * b^n) : b = 1 :=
have bpos : b > 0, from pos_of_ne_zero
(suppose b = 0,
have a^n = 0,
by rewrite [H, this, zero_pow npos],
assert a = 0,
from eq_zero_of_pow_eq_zero this,
show false,
from ne_of_lt `0 < a` this⁻¹),
have H₁ : ∀ p, prime p → ¬ p b, from
take p,
suppose prime p,
suppose p b,
assert p b^n,
from dvd_pow_of_dvd_of_pos `p b` `n > 0`,
have p a^n,
by rewrite H; apply dvd_mul_of_dvd_right this,
have p a,
from dvd_of_prime_of_dvd_pow `prime p` this,
have ¬ coprime a b,
from not_coprime_of_dvd_of_dvd (gt_one_of_prime `prime p`) `p a` `p b`,
show false,
from this `coprime a b`,
have blt2 : b < 2,
from by_contradiction
(suppose ¬ b < 2,
have b ≥ 2,
from le_of_not_gt this,
obtain p [primep pdvdb],
from exists_prime_and_dvd this,
show false,
from H₁ p primep pdvdb),
show b = 1,
from (le.antisymm (le_of_lt_succ blt2) (succ_le_of_lt bpos))
end
/-
Here we state this in terms of the rationals, . The main difficulty is casting between , ,
and .
-/
section
open rat int nat decidable
theorem denom_eq_one_of_pow_eq {q : } {n : } {c : } (npos : n > 0) (H : q^n = c) :
denom q = 1 :=
let a := num q, b := denom q in
have b ≠ 0,
from ne_of_gt (denom_pos q),
have bnz : b ≠ (0 : ),
from assume H, `b ≠ 0` (of_int.inj H),
have bnnz : ((b : rat)^n ≠ 0),
from assume bneqz, bnz (eq_zero_of_pow_eq_zero bneqz),
have a^n /[rat] b^n = c,
using bnz, begin rewrite [*of_int_pow, -div_pow, -eq_num_div_denom, -H] end,
have (a^n : rat) = c *[rat] b^n,
from eq.symm (!mul_eq_of_eq_div bnnz this⁻¹),
have a^n = c * b^n, -- int version
using this, by rewrite [-of_int_pow at this, -of_int_mul at this]; exact of_int.inj this,
have (abs a)^n = abs c * (abs b)^n,
using this, by rewrite [-abs_pow, this, abs_mul, abs_pow],
have H₁ : (nat_abs a)^n = nat_abs c * (nat_abs b)^n,
using this,
begin apply int.of_nat.inj, rewrite [int.of_nat_mul, +int.of_nat_pow, +of_nat_nat_abs],
exact this end,
have H₂ : nat.coprime (nat_abs a) (nat_abs b),
from of_nat.inj !coprime_num_denom,
have nat_abs b = 1, from
by_cases
(suppose q = 0,
by rewrite this)
(suppose qne0 : q ≠ 0,
using H₁ H₂, begin
have ane0 : a ≠ 0, from
suppose aeq0 : a = 0,
have qeq0 : q = 0,
by rewrite [eq_num_div_denom, aeq0, of_int_zero, zero_div],
show false,
from qne0 qeq0,
have nat_abs a ≠ 0, from
suppose nat_abs a = 0,
have aeq0 : a = 0,
from eq_zero_of_nat_abs_eq_zero this,
show false, from ane0 aeq0,
show nat_abs b = 1, from (root_irrational npos (pos_of_ne_zero this) H₂ H₁)
end),
show b = 1,
using this, begin rewrite [-of_nat_nat_abs_of_nonneg (le_of_lt !denom_pos), this] end
theorem eq_num_pow_of_pow_eq {q : } {n : } {c : } (npos : n > 0) (H : q^n = c) :
c = (num q)^n :=
have denom q = 1,
from denom_eq_one_of_pow_eq npos H,
have of_int c = of_int ((num q)^n), using this,
by rewrite [-H, eq_num_div_denom q at {1}, this, of_int_one, div_one, of_int_pow],
show c = (num q)^n , from of_int.inj this
end
/- As a corollary, for n > 1, the nth root of a prime is irrational. -/
section
open nat
theorem not_eq_pow_of_prime {p n : } (a : ) (ngt1 : n > 1) (primep : prime p) : p ≠ a^n :=
assume peq : p = a^n,
have npos : n > 0,
from lt.trans dec_trivial ngt1,
have pnez : p ≠ 0, from
(suppose p = 0,
show false,
by let H := (pos_of_prime primep); rewrite this at H; exfalso; exact !lt.irrefl H),
assert agtz : a > 0, from pos_of_ne_zero
(suppose a = 0,
show false, using npos pnez, by revert peq; rewrite [this, zero_pow npos]; exact pnez),
have n * mult p a = 1, from calc
n * mult p a = mult p (a^n) : begin rewrite [mult_pow n agtz primep] end
... = mult p p : peq
... = 1 : mult_self (gt_one_of_prime primep),
have n 1,
from dvd_of_mul_right_eq this,
have n = 1,
from eq_one_of_dvd_one this,
show false, using this,
by rewrite this at ngt1; exact !lt.irrefl ngt1
open int rat
theorem root_prime_irrational {p n : } {q : } (qnonneg : q ≥ 0) (ngt1 : n > 1)
(primep : prime p) :
q^n ≠ p :=
have numq : num q ≥ 0, from num_nonneg_of_nonneg qnonneg,
have npos : n > 0, from lt.trans dec_trivial ngt1,
suppose q^n = p,
have p = (num q)^n, from eq_num_pow_of_pow_eq npos this,
have p = (nat_abs (num q))^n, using this numq,
by apply of_nat.inj; rewrite [this, of_nat_pow, of_nat_nat_abs_of_nonneg numq],
show false, from not_eq_pow_of_prime _ ngt1 primep this
end
/-
Thaetetus, who lives in the fourth century BC, is said to have proved the irrationality of square
roots up to seventeen. In Chapter 4 of /Why Prove it Again/, John Dawson notes that Thaetetus may
have used an approach similar to the one below. (See data/nat/gcd.lean for the key theorem,
"div_gcd_eq_div_gcd".)
-/
section
open int
example {a b c : } (co : coprime a b) (apos : a > 0) (bpos : b > 0)
(H : a * a = c * (b * b)) :
b = 1 :=
assert H₁ : gcd (c * b) a = gcd c a,
from gcd_mul_right_cancel_of_coprime _ (coprime_swap co),
have a * a = c * b * b,
by rewrite -mul.assoc at H; apply H,
have a / (gcd a b) = c * b / gcd (c * b) a,
from div_gcd_eq_div_gcd this bpos apos,
have a = c * b / gcd c a, using this,
by revert this; rewrite [↑coprime at co, co, int.div_one, H₁]; intros; assumption,
have a = b * (c / gcd c a), using this,
by revert this; rewrite [mul.comm, !int.mul_div_assoc !gcd_dvd_left]; intros; assumption,
have b a,
from dvd_of_mul_right_eq this⁻¹,
have b gcd a b,
from dvd_gcd this !dvd.refl,
have b 1, using this,
by rewrite [↑coprime at co, co at this]; apply this,
show b = 1,
from eq_one_of_dvd_one (le_of_lt bpos) this
end