14-240/Tutorial-October7

DBN: Classes: 2014-15: 14-240 / Navigation Welcome to Math 240! (additions to this web site no longer count towards good deed points) # Week of... Notes and Links 1 Sep 8 About This Class, What is this class about? (PDF, HTML), Monday, Wednesday 2 Sep 15 HW1, Monday, Wednesday, TheComplexField.pdf,HW1_solutions.pdf 3 Sep 22 HW2, Class Photo, Monday, Wednesday, HW2_solutions.pdf 4 Sep 29 HW3, Wednesday, Tutorial, HW3_solutions.pdf 5 Oct 6 HW4, Monday, Wednesday, Tutorial, HW4_solutions.pdf 6 Oct 13 No Monday class (Thanksgiving), Wednesday, Tutorial 7 Oct 20 HW5, Term Test at tutorials on Tuesday, Wednesday 8 Oct 27 HW6, Monday, Why LinAlg?, Wednesday, Tutorial 9 Nov 3 Monday is the last day to drop this class, HW7, Monday, Wednesday, Tutorial 10 Nov 10 HW8, Monday, Tutorial 11 Nov 17 Monday-Tuesday is UofT November break 12 Nov 24 HW9 13 Dec 1 Wednesday is a "makeup Monday"! End-of-Course Schedule, Tutorial F Dec 8 The Final Exam Register of Good Deeds Add your name / see who's in!

Boris

Subtle Errors in Proofs

Check out these proofs:

Proof 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 }[/math], [math]\displaystyle{ y \in W_2 }[/math].

Then [math]\displaystyle{ x, y \in W_1 \cup W_2 }[/math].

Then [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.

Proof 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].

We show that [math]\displaystyle{ V }[/math] is not commutative.

Let [math]\displaystyle{ (a_1, a_2) = (0, 0) }[/math].

Then [math]\displaystyle{ (0, 0) + (b_1, b_2) = (2b_1, 3b_2) \neq (b_1, b_2) }[/math].

Then [math]\displaystyle{ V }[/math] is not commutative.

Then [math]\displaystyle{ V }[/math] is not a vector space. Q.E.D.

Can you spot the subtle error in each?

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

(1) Let [math]\displaystyle{ x \in W_1, y \in W_2 }[/math]. [Many lines] Then [math]\displaystyle{ x \in W_2 \or y \in W_1 }[/math].

(2) Then [math]\displaystyle{ W_1 \subset W_2 \or W_2 \subset W_1 }[/math].

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

(1) [math]\displaystyle{ \forall x \in W_1, \forall y \in W_2, (x \in W_2 \or y \in W_1) }[/math].

(2) [math]\displaystyle{ ( \forall x \in W_1, x \in W_2) \or ( \forall y \in W_2, y \in W_1) }[/math].

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 [math]\displaystyle{ (0, 0) }[/math] is not the additive identity. For Proof 2 to be correct, either plug in a vector that is not [math]\displaystyle{ (0, 0) }[/math] or show that [math]\displaystyle{ (0, 0) }[/math] is the additive identity by some other means, which introduces a contradiction.