Difference between revisions of "14-240/Tutorial-October7"

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(Proof 2)
(Proof 1)
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=====Proof 1=====
 
=====Proof 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
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Let <math>W_1</math>, <math>W_2</math> be subspaces of a vector space <math>V</math>.
  
<math>\implies W_1 \subset W_2 \or W_2 \subset W_1</math>.
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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.

Revision as of 21:17, 15 October 2014

Contents

Boris

Subtle Errors in Proofs

Check out these proofs:

Proof 1

Let W_1, W_2 be subspaces of a vector space V.

We show that W_1 \cup W_2 is a subspace \implies W_1 \subset W_2 \or W_2 \subset W_1.

Assume that W_1 \cup W_2 is a subspace.
Let x \in W_1, y \in W_2.
Then x, y \in W_1 \cup W_2.
Then x + y \in W_1 \cup W_2.
Then x + y \in W_1 \or x + y \in W_2.
Case 1: x + y \in W_1:
Since x \in W_1 and W_1 has additive inverses, then (-x) \in W_1.
Then (x+y)+(-x)=y \in W_1.
Case 2: x + y \in W_2:
Since y \in W_2 and W_2 has additive inverses, then (-y) \in W_2.
Then (x+y)+(-y)=x \in W_2.
Then x \in W_2 \or y \in W_1.
Then W_1 \subset W_2 \or W_2 \subset W_1. Q.E.D.
Proof 2

Let V=\{(a_1, a_2):a_1, a_2 \in R\}. Then \forall (a_1, a_2), (b_1, b_2) \in V, \forall c \in R, define

(a_1, a_2) + (b_1, b_2) = (a_1 + 2b_1, a_2 + 3b_2) and c(a_1, a_2)=(ca_1, ca_2). We show that V is not a vector

space over R.

We show that V is not commutative.
Let (a_1, a_2) = (0, 0).
Then (0, 0) + (b_1, b_2) = (2b_1, 3b_2) \neq (b_1, b_2).
Then V is not commutative.
Then V is not a vector space. Q.E.D.


Can you spot the subtle error in each?


In Proof 1, the equivalence of the "let" statement to the second last line and the last line is not obvious:

(1) Let x \in W_1, y \in W_2. [Many lines] Then x \in W_2 \or y \in W_1.

(2) Then W_1 \subset W_2 \or W_2 \subset W_1.


Rewrite sentences (1) and (2) into a form that is easier to compare:

(1) \forall x \in W_1, \forall y \in W_2, (x \in W_2 \or y \in W_1).

(2) ( \forall x \in W_1, x \in W_2) \or ( \forall y \in W_2, y \in W_1).


For Proof 1 to be correct, we must show that sentences (1) and (2) are equivalent. Alternatively, alter the structure of Proof 1 into a proof by contradiction or into proof in which you assume that one of the conditions in the disjunction is satisfied.


In Proof 2, the only thing that is shown is that (0, 0) is not the additive identity. For Proof 2 to be correct, either plug in a vector that is not (0, 0) or show that (0, 0) is the additive identity by some other means, which introduces a contradiction.

Nikita