(* Title: Roots of real quadratics
Author: Tim Makarios <tjm1983 at gmail.com>, 2012
Originally from the AFP entry Tarskis_Geometry
*)
section "Roots of real quadratics"
theory Quadratic_Discriminant
imports Complex_Main
begin
definition discrim :: "[real,real,real] \<Rightarrow> real" where
"discrim a b c \<equiv> b\<^sup>2 - 4 * a * c"
lemma complete_square:
fixes a b c x :: "real"
assumes "a \<noteq> 0"
shows "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow> (2 * a * x + b)\<^sup>2 = discrim a b c"
proof -
have "4 * a\<^sup>2 * x\<^sup>2 + 4 * a * b * x + 4 * a * c = 4 * a * (a * x\<^sup>2 + b * x + c)"
by (simp add: algebra_simps power2_eq_square)
with \<open>a \<noteq> 0\<close>
have "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow> 4 * a\<^sup>2 * x\<^sup>2 + 4 * a * b * x + 4 * a * c = 0"
by simp
thus "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow> (2 * a * x + b)\<^sup>2 = discrim a b c"
unfolding discrim_def
by (simp add: power2_eq_square algebra_simps)
qed
lemma discriminant_negative:
fixes a b c x :: real
assumes "a \<noteq> 0"
and "discrim a b c < 0"
shows "a * x\<^sup>2 + b * x + c \<noteq> 0"
proof -
have "(2 * a * x + b)\<^sup>2 \<ge> 0" by simp
with \<open>discrim a b c < 0\<close> have "(2 * a * x + b)\<^sup>2 \<noteq> discrim a b c" by arith
with complete_square and \<open>a \<noteq> 0\<close> show "a * x\<^sup>2 + b * x + c \<noteq> 0" by simp
qed
lemma plus_or_minus_sqrt:
fixes x y :: real
assumes "y \<ge> 0"
shows "x\<^sup>2 = y \<longleftrightarrow> x = sqrt y \<or> x = - sqrt y"
proof
assume "x\<^sup>2 = y"
hence "sqrt (x\<^sup>2) = sqrt y" by simp
hence "sqrt y = \<bar>x\<bar>" by simp
thus "x = sqrt y \<or> x = - sqrt y" by auto
next
assume "x = sqrt y \<or> x = - sqrt y"
hence "x\<^sup>2 = (sqrt y)\<^sup>2 \<or> x\<^sup>2 = (- sqrt y)\<^sup>2" by auto
with \<open>y \<ge> 0\<close> show "x\<^sup>2 = y" by simp
qed
lemma divide_non_zero:
fixes x y z :: real
assumes "x \<noteq> 0"
shows "x * y = z \<longleftrightarrow> y = z / x"
proof
assume "x * y = z"
with \<open>x \<noteq> 0\<close> show "y = z / x" by (simp add: field_simps)
next
assume "y = z / x"
with \<open>x \<noteq> 0\<close> show "x * y = z" by simp
qed
lemma discriminant_nonneg:
fixes a b c x :: real
assumes "a \<noteq> 0"
and "discrim a b c \<ge> 0"
shows "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow>
x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a)"
proof -
from complete_square and plus_or_minus_sqrt and assms
have "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow>
(2 * a) * x + b = sqrt (discrim a b c) \<or>
(2 * a) * x + b = - sqrt (discrim a b c)"
by simp
also have "\<dots> \<longleftrightarrow> (2 * a) * x = (-b + sqrt (discrim a b c)) \<or>
(2 * a) * x = (-b - sqrt (discrim a b c))"
by auto
also from \<open>a \<noteq> 0\<close> and divide_non_zero [of "2 * a" x]
have "\<dots> \<longleftrightarrow> x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a)"
by simp
finally show "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow>
x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a)" .
qed
lemma discriminant_zero:
fixes a b c x :: real
assumes "a \<noteq> 0"
and "discrim a b c = 0"
shows "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow> x = -b / (2 * a)"
using discriminant_nonneg and assms
by simp
theorem discriminant_iff:
fixes a b c x :: real
assumes "a \<noteq> 0"
shows "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow>
discrim a b c \<ge> 0 \<and>
(x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a))"
proof
assume "a * x\<^sup>2 + b * x + c = 0"
with discriminant_negative and \<open>a \<noteq> 0\<close> have "\<not>(discrim a b c < 0)" by auto
hence "discrim a b c \<ge> 0" by simp
with discriminant_nonneg and \<open>a * x\<^sup>2 + b * x + c = 0\<close> and \<open>a \<noteq> 0\<close>
have "x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a)"
by simp
with \<open>discrim a b c \<ge> 0\<close>
show "discrim a b c \<ge> 0 \<and>
(x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a))" ..
next
assume "discrim a b c \<ge> 0 \<and>
(x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a))"
hence "discrim a b c \<ge> 0" and
"x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a)"
by simp_all
with discriminant_nonneg and \<open>a \<noteq> 0\<close> show "a * x\<^sup>2 + b * x + c = 0" by simp
qed
lemma discriminant_nonneg_ex:
fixes a b c :: real
assumes "a \<noteq> 0"
and "discrim a b c \<ge> 0"
shows "\<exists> x. a * x\<^sup>2 + b * x + c = 0"
using discriminant_nonneg and assms
by auto
lemma discriminant_pos_ex:
fixes a b c :: real
assumes "a \<noteq> 0"
and "discrim a b c > 0"
shows "\<exists> x y. x \<noteq> y \<and> a * x\<^sup>2 + b * x + c = 0 \<and> a * y\<^sup>2 + b * y + c = 0"
proof -
let ?x = "(-b + sqrt (discrim a b c)) / (2 * a)"
let ?y = "(-b - sqrt (discrim a b c)) / (2 * a)"
from \<open>discrim a b c > 0\<close> have "sqrt (discrim a b c) \<noteq> 0" by simp
hence "sqrt (discrim a b c) \<noteq> - sqrt (discrim a b c)" by arith
with \<open>a \<noteq> 0\<close> have "?x \<noteq> ?y" by simp
moreover
from discriminant_nonneg [of a b c ?x]
and discriminant_nonneg [of a b c ?y]
and assms
have "a * ?x\<^sup>2 + b * ?x + c = 0" and "a * ?y\<^sup>2 + b * ?y + c = 0" by simp_all
ultimately
show "\<exists> x y. x \<noteq> y \<and> a * x\<^sup>2 + b * x + c = 0 \<and> a * y\<^sup>2 + b * y + c = 0" by blast
qed
lemma discriminant_pos_distinct:
fixes a b c x :: real
assumes "a \<noteq> 0" and "discrim a b c > 0"
shows "\<exists> y. x \<noteq> y \<and> a * y\<^sup>2 + b * y + c = 0"
proof -
from discriminant_pos_ex and \<open>a \<noteq> 0\<close> and \<open>discrim a b c > 0\<close>
obtain w and z where "w \<noteq> z"
and "a * w\<^sup>2 + b * w + c = 0" and "a * z\<^sup>2 + b * z + c = 0"
by blast
show "\<exists> y. x \<noteq> y \<and> a * y\<^sup>2 + b * y + c = 0"
proof cases
assume "x = w"
with \<open>w \<noteq> z\<close> have "x \<noteq> z" by simp
with \<open>a * z\<^sup>2 + b * z + c = 0\<close>
show "\<exists> y. x \<noteq> y \<and> a * y\<^sup>2 + b * y + c = 0" by auto
next
assume "x \<noteq> w"
with \<open>a * w\<^sup>2 + b * w + c = 0\<close>
show "\<exists> y. x \<noteq> y \<and> a * y\<^sup>2 + b * y + c = 0" by auto
qed
qed
end