14-240/Tutorial-October7: Difference between revisions
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(1) Let <math>W_1</math>, <math>W_2</math> be subspaces of a vector space <math>V</math>. We show that <math>W_1 \cup W_2</math> is a subspace <math>\implies W_1 \subset W_2 \or W_2 \subset W_1</math>. |
(1) Let <math>W_1</math>, <math>W_2</math> be subspaces of a vector space <math>V</math>. We show that <math>W_1 \cup W_2</math> is a subspace <math>\implies W_1 \subset W_2 \or W_2 \subset W_1</math>. |
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:Assume that <math>W_1 \cup W_2</math> is a subspace. |
:Assume that <math>W_1 \cup W_2</math> is a subspace. |
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:Let <math>x \in W_1, y \in W_2</math>. |
:Let <math>x \in W_1, y \in W_2</math>. |
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:Then <math>x, y \in W_1 \cup W_2</math> and <math>x + y \in W_1 \cup W_2</math>. |
:Then <math>x, y \in W_1 \cup W_2</math> and <math>x + y \in W_1 \cup W_2</math>. |
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:Then <math>x + y \in W_1 \or x + y \in W_2</math>. |
:Then <math>x + y \in W_1 \or x + y \in W_2</math>. |
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:Case 1: <math>x + y \in W_1</math>: |
:Case 1: <math>x + y \in W_1</math>: |
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::Since <math>x \in W_1</math> and <math>W_1</math> has additive inverses, then <math>(-x) \in W_1</math>. |
::Since <math>x \in W_1</math> and <math>W_1</math> has additive inverses, then <math>(-x) \in W_1</math>. |
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::Then <math>(x+y)+(-x)=y \in W_1</math>. |
::Then <math>(x+y)+(-x)=y \in W_1</math>. |
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:Case 2: <math>x + y \in W_2</math>: |
:Case 2: <math>x + y \in W_2</math>: |
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::Since <math>y \in W_2</math> and <math>W_2</math> has additive inverses, then <math>(-y) \in W_2</math>. |
::Since <math>y \in W_2</math> and <math>W_2</math> has additive inverses, then <math>(-y) \in W_2</math>. |
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::Then <math>(x+y)+(-y)=x \in W_2</math>. |
::Then <math>(x+y)+(-y)=x \in W_2</math>. |
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:Then <math>x \in W_2 \or y \in W_1</math>. |
:Then <math>x \in W_2 \or y \in W_1</math>. |
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:Then <math>W_1 \subset W_2 \or W_2 \subset W_1</math>. ''Q.E.D.'' |
:Then <math>W_1 \subset W_2 \or W_2 \subset W_1</math>. ''Q.E.D.'' |
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Revision as of 23:37, 11 October 2014
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Boris
Subtle Problems in Proofs
Check out these proofs:
(1) Let [math]\displaystyle{ W_1 }[/math], [math]\displaystyle{ W_2 }[/math] be subspaces of a vector space [math]\displaystyle{ V }[/math]. We show that [math]\displaystyle{ W_1 \cup W_2 }[/math] is a subspace [math]\displaystyle{ \implies W_1 \subset W_2 \or W_2 \subset W_1 }[/math].
- Assume that [math]\displaystyle{ W_1 \cup W_2 }[/math] is a subspace.
- Let [math]\displaystyle{ x \in W_1, y \in W_2 }[/math].
- Then [math]\displaystyle{ x, y \in W_1 \cup W_2 }[/math] and [math]\displaystyle{ x + y \in W_1 \cup W_2 }[/math].
- Then [math]\displaystyle{ x + y \in W_1 \or x + y \in W_2 }[/math].
- Case 1: [math]\displaystyle{ x + y \in W_1 }[/math]:
- Since [math]\displaystyle{ x \in W_1 }[/math] and [math]\displaystyle{ W_1 }[/math] has additive inverses, then [math]\displaystyle{ (-x) \in W_1 }[/math].
- Then [math]\displaystyle{ (x+y)+(-x)=y \in W_1 }[/math].
- Case 2: [math]\displaystyle{ x + y \in W_2 }[/math]:
- Since [math]\displaystyle{ y \in W_2 }[/math] and [math]\displaystyle{ W_2 }[/math] has additive inverses, then [math]\displaystyle{ (-y) \in W_2 }[/math].
- Then [math]\displaystyle{ (x+y)+(-y)=x \in W_2 }[/math].
- Then [math]\displaystyle{ x \in W_2 \or y \in W_1 }[/math].
- Then [math]\displaystyle{ W_1 \subset W_2 \or W_2 \subset W_1 }[/math]. Q.E.D.
(2) Let [math]\displaystyle{ V=\{(a_1, a_2):a_1, a_2 \in R\} }[/math]. Then [math]\displaystyle{ \forall (a_1, a_2), (b_1, b_2) \in V, \forall c \in R }[/math], define [math]\displaystyle{ (a_1, a_2) + (b_1, b_2) = (a_1 + 2b_1, a_2 + 3b_2) }[/math] and [math]\displaystyle{ c(a_1, a_2)=(ca_1, ca_2) }[/math]. We show that [math]\displaystyle{ V }[/math] is not a vector space over [math]\displaystyle{ R }[/math].
