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, y \in W_1 \cup W_2 }[/math].

Then [math]\displaystyle{ x + y \in W_1 \cup W_2 }[/math] and [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 \cup W_2 }[/math] and [math]\displaystyle{ W_1 \cup W_2 }[/math] has additive inverses, then [math]\displaystyle{ (-x) \in W_1 \cup W_2 }[/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_1 \cup W_2 }[/math] and [math]\displaystyle{ W_1 \cup W_2 }[/math] has additive inverses, then [math]\displaystyle{ (-y) \in W_1 \cup 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) = (b_1, b_2) + (0, 0) }[/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?

Error in Proof 1

The error in the proof is the logical equivalence of these two statements.

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