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>.
:Assume that <math>W_1 \cup W_2</math> is a subspace.
:Assume that <math>W_1 \cup W_2</math> is a subspace.

:Let <math>x \in W_1, y \in W_2</math>.
:Let <math>x \in W_1, y \in 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>.
: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 \or x + y \in W_2</math>.
:Then <math>x + y \in W_1 \or x + y \in W_2</math>.

:Case 1: <math>x + y \in W_1</math>:
:Case 1: <math>x + y \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>.
::Since <math>x \in W_1</math> and <math>W_1</math> has additive inverses, then <math>(-x) \in W_1</math>.

::Then <math>(x+y)+(-x)=y \in W_1</math>.
::Then <math>(x+y)+(-x)=y \in W_1</math>.

:Case 2: <math>x + y \in W_2</math>:
:Case 2: <math>x + 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>.
::Since <math>y \in W_2</math> and <math>W_2</math> has additive inverses, then <math>(-y) \in W_2</math>.

::Then <math>(x+y)+(-y)=x \in W_2</math>.
::Then <math>(x+y)+(-y)=x \in W_2</math>.

:Then <math>x \in W_2 \or y \in W_1</math>.
:Then <math>x \in W_2 \or y \in W_1</math>.

: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.''



Revision as of 23:37, 11 October 2014

Boris

Subtle Problems in Proofs

Check out these proofs:

(1) Let , be subspaces of a vector space . We show that is a subspace .

Assume that is a subspace.
Let .
Then and .
Then .
Case 1: :
Since and has additive inverses, then .
Then .
Case 2: :
Since and has additive inverses, then .
Then .
Then .
Then . Q.E.D.


(2) Let . Then , define and . We show that is not a vector space over .

Nikita