(* Title: HOL/Hyperreal/ex/Sqrt.thy
ID: $Id: Sqrt.thy,v 1.2 2002/03/06 17:16:48 wenzelm Exp $
Author: Markus Wenzel, TU Muenchen
License: GPL (GNU GENERAL PUBLIC LICENSE)
*)
header {* Square roots of primes are irrational *}
theory Sqrt = Primes + Hyperreal:
subsection {* Preliminaries *}
text {*
The set of rational numbers, including the key representation
theorem.
*}
constdefs
rationals :: "real set" ("\")
"\ \ {x. \m n. n \ 0 \ \x\ = real (m::nat) / real (n::nat)}"
theorem rationals_rep: "x \ \ \
\m n. n \ 0 \ \x\ = real m / real n \ gcd (m, n) = 1"
proof -
assume "x \ \"
then obtain m n :: nat where
n: "n \ 0" and x_rat: "\x\ = real m / real n"
by (unfold rationals_def) blast
let ?gcd = "gcd (m, n)"
from n have gcd: "?gcd \ 0" by (simp add: gcd_zero)
let ?k = "m div ?gcd"
let ?l = "n div ?gcd"
let ?gcd' = "gcd (?k, ?l)"
have "?gcd dvd m" .. then have gcd_k: "?gcd * ?k = m"
by (rule dvd_mult_div_cancel)
have "?gcd dvd n" .. then have gcd_l: "?gcd * ?l = n"
by (rule dvd_mult_div_cancel)
from n and gcd_l have "?l \ 0"
by (auto iff del: neq0_conv)
moreover
have "\x\ = real ?k / real ?l"
proof -
from gcd have "real ?k / real ?l =
real (?gcd * ?k) / real (?gcd * ?l)"
by (simp add: real_mult_div_cancel1)
also from gcd_k and gcd_l have "\ = real m / real n" by simp
also from x_rat have "\ = \x\" ..
finally show ?thesis ..
qed
moreover
have "?gcd' = 1"
proof -
have "?gcd * ?gcd' = gcd (?gcd * ?k, ?gcd * ?l)"
by (rule gcd_mult_distrib2)
with gcd_k gcd_l have "?gcd * ?gcd' = ?gcd" by simp
with gcd show ?thesis by simp
qed
ultimately show ?thesis by blast
qed
lemma [elim?]: "r \ \ \
(\m n. n \ 0 \ \r\ = real m / real n \ gcd (m, n) = 1 \ C)
\ C"
using rationals_rep by blast
subsection {* Main theorem *}
text {*
The square root of any prime number (including @{text 2}) is
irrational.
*}
theorem sqrt_prime_irrational: "p \ prime \ sqrt (real p) \ \"
proof
assume p_prime: "p \ prime"
then have p: "1 < p" by (simp add: prime_def)
assume "sqrt (real p) \ \"
then obtain m n where
n: "n \ 0" and sqrt_rat: "\sqrt (real p)\ = real m / real n"
and gcd: "gcd (m, n) = 1" ..
have eq: "m\ = p * n\"
proof -
from n and sqrt_rat have "real m = \sqrt (real p)\ * real n" by simp
then have "real (m\) = (sqrt (real p))\ * real (n\)"
by (auto simp add: power_two real_power_two)
also have "(sqrt (real p))\ = real p" by simp
also have "\ * real (n\) = real (p * n\)" by simp
finally show ?thesis ..
qed
have "p dvd m \ p dvd n"
proof
from eq have "p dvd m\" ..
with p_prime show "p dvd m" by (rule prime_dvd_power_two)
then obtain k where "m = p * k" ..
with eq have "p * n\ = p\ * k\" by (auto simp add: power_two mult_ac)
with p have "n\ = p * k\" by (simp add: power_two)
then have "p dvd n\" ..
with p_prime show "p dvd n" by (rule prime_dvd_power_two)
qed
then have "p dvd gcd (m, n)" ..
with gcd have "p dvd 1" by simp
then have "p \ 1" by (simp add: dvd_imp_le)
with p show False by simp
qed
corollary "sqrt (real (2::nat)) \ \"
by (rule sqrt_prime_irrational) (rule two_is_prime)
subsection {* Variations *}
text {*
Here is an alternative version of the main proof, using mostly
linear forward-reasoning. While this results in less top-down
structure, it is probably closer to proofs seen in mathematics.
*}
theorem "p \ prime \ sqrt (real p) \ \"
proof
assume p_prime: "p \ prime"
then have p: "1 < p" by (simp add: prime_def)
assume "sqrt (real p) \ \"
then obtain m n where
n: "n \ 0" and sqrt_rat: "\sqrt (real p)\ = real m / real n"
and gcd: "gcd (m, n) = 1" ..
from n and sqrt_rat have "real m = \sqrt (real p)\ * real n" by simp
then have "real (m\) = (sqrt (real p))\ * real (n\)"
by (auto simp add: power_two real_power_two)
also have "(sqrt (real p))\ = real p" by simp
also have "\ * real (n\) = real (p * n\)" by simp
finally have eq: "m\ = p * n\" ..
then have "p dvd m\" ..
with p_prime have dvd_m: "p dvd m" by (rule prime_dvd_power_two)
then obtain k where "m = p * k" ..
with eq have "p * n\ = p\ * k\" by (auto simp add: power_two mult_ac)
with p have "n\ = p * k\" by (simp add: power_two)
then have "p dvd n\" ..
with p_prime have "p dvd n" by (rule prime_dvd_power_two)
with dvd_m have "p dvd gcd (m, n)" by (rule gcd_greatest)
with gcd have "p dvd 1" by simp
then have "p \ 1" by (simp add: dvd_imp_le)
with p show False by simp
qed
end