Download Lines in n-Dimensional Euclidean Spaces

FORMALIZED MATHEMATICS
11
4
Volume
,
Number ,
University of Białystok
2003
Lines in n-Dimensional Euclidean Spaces
Akihiro Kubo
Shinshu University
Nagano
Summary. In this paper, we define the line of n-dimensional Euclidean
space and we introduce basic properties of affine space on this space. Next, we
define the inner product of elements of this space. At the end, we introduce
orthogonality of lines of this space.
MML Identifier: EUCLID 4.
The papers [13], [4], [15], [2], [12], [8], [5], [11], [10], [3], [6], [1], [14], [7], and [9]
provide the terminology and notation for this paper.
We adopt the following rules: a, b, l1 are real numbers, n is a natural number,
and x, x1 , x2 , y1 , y2 are elements of Rn .
Next we state several propositions:
(1) 0 · x + x = x and x + h0, . . . , 0i = x.
| {z }
n
(2) a · h0, . . . , 0i = h0, . . . , 0i.
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n
n
(3) 1 · x = x and 0 · x = h0, . . . , 0i.
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(4) (a · b) · x = a · (b · x).
(5) If a · x = h0, . . . , 0i, then a = 0 or x = h0, . . . , 0i.
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| {z }
n
n
(6) a · (x1 + x2 ) = a · x1 + a · x2 .
(7) (a + b) · x = a · x + b · x.
(8) If a · x1 = a · x2 , then a = 0 or x1 = x2 .
Let us consider n and let x1 , x2 be elements of Rn . The functor Line(x1 , x2 )
yields a subset of Rn and is defined by:
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2003 University of Białystok
ISSN 1426–2630
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akihiro kubo
(Def. 1) Line(x1 , x2 ) = {(1 − l1 ) · x1 + l1 · x2 }.
Let us consider n and let x1 , x2 be elements of Rn . Observe that Line(x1 , x2 )
is non empty.
The following proposition is true
(9) Line(x1 , x2 ) = Line(x2 , x1 ).
Let us consider n and let x1 , x2 be elements of Rn . Let us observe that the
functor Line(x1 , x2 ) is commutative.
One can prove the following propositions:
(10) x1 ∈ Line(x1 , x2 ) and x2 ∈ Line(x1 , x2 ).
(11) If y1 ∈ Line(x1 , x2 ) and y2 ∈ Line(x1 , x2 ), then Line(y1 , y2 ) ⊆
Line(x1 , x2 ).
(12) If y1 ∈ Line(x1 , x2 ) and y2 ∈ Line(x1 , x2 ) and y1 6= y2 , then
Line(x1 , x2 ) ⊆ Line(y1 , y2 ).
Let us consider n and let A be a subset of Rn . We say that A is line if and
only if:
(Def. 2) There exist x1 , x2 such that x1 6= x2 and A = Line(x1 , x2 ).
We introduce A is a line as a synonym of A is line.
Next we state three propositions:
(13) Let A, C be subsets of Rn and given x1 , x2 . Suppose A is a line and C
is a line and x1 ∈ A and x2 ∈ A and x1 ∈ C and x2 ∈ C. Then x1 = x2 or
A = C.
(14) For every subset A of Rn such that A is a line there exist x1 , x2 such
that x1 ∈ A and x2 ∈ A and x1 6= x2 .
(15) For every subset A of Rn such that A is a line there exists x2 such that
x1 6= x2 and x2 ∈ A.
Let us consider n and let x be an element of Rn . The functor Rn2Fin(x)
yielding a finite sequence of elements of R is defined by:
(Def. 3) Rn2Fin(x) = x.
Let us consider n and let x be an element of Rn . The functor |x| yields a
real number and is defined as follows:
(Def. 4) |x| = | Rn2Fin(x)|.
Let us consider n and let x1 , x2 be elements of Rn . The functor |(x1 , x2 )|
yielding a real number is defined by:
(Def. 5) |(x1 , x2 )| = |(Rn2Fin(x1 ), Rn2Fin(x2 ))|.
Let us observe that the functor |(x1 , x2 )| is commutative.
We now state a number of propositions:
(16) For all elements x1 , x2 of Rn holds |(x1 , x2 )| = 41 ·(|x1 +x2 |2 −|x1 −x2 |2 ).
(17) For every element x of Rn holds |(x, x)| ­ 0.
(18) For every element x of Rn holds |x|2 = |(x, x)|.
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(19)
(20)
(21)
(22)
For
For
For
For
every
every
every
every
element
element
element
element
of
of
of
of
Rn
Rn
Rn
Rn
(23) For every element x of
Rn
x
x
x
x
holds
holds
holds
holds
0 ¬ |x|.
p
|x| = |(x, x)|.
|(x, x)| = 0 iff |x| = 0.
|(x, x)| = 0 iff x = h0, . . . , 0i.
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holds |(x, h0, . . . , 0i)| = 0.
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(24) For every element x of Rn holds |(h0, . . . , 0i, x)| = 0.
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(25) For all elements x1 , x2 , x3 of Rn holds |(x1 + x2 , x3 )| = |(x1 , x3 )| +
|(x2 , x3 )|.
(26) For all elements x1 , x2 of Rn and for every real number a holds |(a ·
x1 , x2 )| = a · |(x1 , x2 )|.
(27) For all elements x1 , x2 of Rn and for every real number a holds |(x1 , a ·
x2 )| = a · |(x1 , x2 )|.
(28) For all elements x1 , x2 of Rn holds |(−x1 , x2 )| = −|(x1 , x2 )|.
(29) For all elements x1 , x2 of Rn holds |(x1 , −x2 )| = −|(x1 , x2 )|.
(30) For all elements x1 , x2 of Rn holds |(−x1 , −x2 )| = |(x1 , x2 )|.
(31) For all elements x1 , x2 , x3 of Rn holds |(x1 − x2 , x3 )| = |(x1 , x3 )| −
|(x2 , x3 )|.
(32) For all real numbers a, b and for all elements x1 , x2 , x3 of Rn holds
|(a · x1 + b · x2 , x3 )| = a · |(x1 , x3 )| + b · |(x2 , x3 )|.
(33) For all elements x1 , y1 , y2 of Rn holds |(x1 , y1 + y2 )| = |(x1 , y1 )| +
|(x1 , y2 )|.
(34) For all elements x1 , y1 , y2 of Rn holds |(x1 , y1 − y2 )| = |(x1 , y1 )| −
|(x1 , y2 )|.
(35) For all elements x1 , x2 , y1 , y2 of Rn holds |(x1 +x2 , y1 +y2 )| = |(x1 , y1 )|+
|(x1 , y2 )| + |(x2 , y1 )| + |(x2 , y2 )|.
(36) For all elements x1 , x2 , y1 , y2 of Rn holds |(x1 −x2 , y1 −y2 )| = (|(x1 , y1 )|−
|(x1 , y2 )| − |(x2 , y1 )|) + |(x2 , y2 )|.
(37) For all elements x, y of Rn holds |(x + y, x + y)| = |(x, x)| + 2 · |(x, y)| +
|(y, y)|.
(38) For all elements x, y of Rn holds |(x − y, x − y)| = (|(x, x)| − 2 · |(x, y)|) +
|(y, y)|.
(39) For all elements x, y of Rn holds |x + y|2 = |x|2 + 2 · |(x, y)| + |y|2 .
(40) For all elements x, y of Rn holds |x − y|2 = (|x|2 − 2 · |(x, y)|) + |y|2 .
(41) For all elements x, y of Rn holds |x + y|2 + |x − y|2 = 2 · (|x|2 + |y|2 ).
(42) For all elements x, y of Rn holds |x + y|2 − |x − y|2 = 4 · |(x, y)|.
(43) For all elements x, y of Rn holds ||(x, y)|| ¬ |x| · |y|.
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akihiro kubo
(44) For all elements x, y of Rn holds |x + y| ¬ |x| + |y|.
Let us consider n and let x1 , x2 be elements of Rn . We say that x1 , x2 are
orthogonal if and only if:
(Def. 6) |(x1 , x2 )| = 0.
Let us note that the predicate x1 , x2 are orthogonal is symmetric.
We now state the proposition
(45) Let R be a subset of R and x1 , x2 , y1 be elements of Rn . Suppose
R = {|y1 − x|; x ranges over elements of Rn : x ∈ Line(x1 , x2 )}. Then there
exists an element y2 of Rn such that y2 ∈ Line(x1 , x2 ) and |y1 −y2 | = inf R
and x1 − x2 , y1 − y2 are orthogonal.
Let us consider n and let p1 , p2 be points of ETn . The functor Line(p1 , p2 )
yielding a subset of ETn is defined by:
(Def. 7) Line(p1 , p2 ) = {(1 − l1 ) · p1 + l1 · p2 }.
Let us consider n and let p1 , p2 be points of ETn . Observe that Line(p1 , p2 ) is
non empty.
In the sequel p1 , p2 , q1 , q2 are points of ETn .
The following proposition is true
(46) Line(p1 , p2 ) = Line(p2 , p1 ).
Let us consider n and let p1 , p2 be points of ETn . Let us observe that the
functor Line(p1 , p2 ) is commutative.
One can prove the following three propositions:
(47) p1 ∈ Line(p1 , p2 ) and p2 ∈ Line(p1 , p2 ).
(48) If q1 ∈ Line(p1 , p2 ) and q2 ∈ Line(p1 , p2 ), then Line(q1 , q2 ) ⊆
Line(p1 , p2 ).
(49) If q1 ∈ Line(p1 , p2 ) and q2 ∈ Line(p1 , p2 ) and q1 6= q2 , then Line(p1 , p2 ) ⊆
Line(q1 , q2 ).
Let us consider n and let A be a subset of ETn . We say that A is line if and
only if:
(Def. 8) There exist p1 , p2 such that p1 6= p2 and A = Line(p1 , p2 ).
We introduce A is a line as a synonym of A is line.
We now state three propositions:
(50) For all subsets A, C of ETn such that A is a line and C is a line and p1 ∈ A
and p2 ∈ A and p1 ∈ C and p2 ∈ C holds p1 = p2 or A = C.
(51) For every subset A of ETn such that A is a line there exist p1 , p2 such
that p1 ∈ A and p2 ∈ A and p1 6= p2 .
(52) For every subset A of ETn such that A is a line there exists p2 such that
p1 6= p2 and p2 ∈ A.
Let us consider n and let p be a point of ETn . The functor TPn2Rn(p) yields
an element of Rn and is defined as follows:
lines in n-dimensional euclidean spaces
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(Def. 9) TPn2Rn(p) = p.
Let us consider n and let p be a point of ETn . The functor |p| yields a real
number and is defined as follows:
(Def. 10) |p| = | TPn2Rn(p)|.
Let us consider n and let p1 , p2 be points of ETn . The functor |(p1 , p2 )| yields
a real number and is defined as follows:
(Def. 11) |(p1 , p2 )| = |(TPn2Rn(p1 ), TPn2Rn(p2 ))|.
Let us observe that the functor |(p1 , p2 )| is commutative.
Let us consider n and let p1 , p2 be points of ETn . We say that p1 , p2 are
orthogonal if and only if:
(Def. 12) |(p1 , p2 )| = 0.
Let us note that the predicate p1 , p2 are orthogonal is symmetric.
Next we state the proposition
(53) Let R be a subset of R and p1 , p2 , q1 be points of ETn . Suppose R =
{|q1 − p|; p ranges over points of ETn : p ∈ Line(p1 , p2 )}. Then there exists a
point q2 of ETn such that q2 ∈ Line(p1 , p2 ) and |q1 − q2 | = inf R and p1 − p2 ,
q1 − q2 are orthogonal.
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Received August 8, 2003