14-240/Tutorial-October7: Difference between revisions

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Check out these proofs:
Check out these proofs:


(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.
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(2) Let <math>V=\{(a_1, a_2):a_1, a_2 \in R\}</math>. Then <math>\forall (a_1, a_2), (b_1, b_2) \in V, \forall c \in R</math>, define <math>(a_1, a_2) + (b_1, b_2) = (a_1 + 2b_1, a_2 + 3b_2)</math> and <math>c(a_1, a_2)=(ca_1, ca_2)</math>. We show that <math>V</math> is not a vector space over <math>R</math>.
(2) Let <math>V=\{(a_1, a_2):a_1, a_2 \in R\}</math>. Then <math>\forall (a_1, a_2), (b_1, b_2) \in V, \forall c \in R</math>, define <math>(a_1, a_2) + (b_1, b_2) = (a_1 + 2b_1, a_2 + 3b_2)</math> and <math>c(a_1, a_2)=(ca_1, ca_2)</math>. We show that <math>V</math> is not a vector

space over <math>R</math>.


==Nikita==
==Nikita==

Revision as of 23:38, 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