Welcome to Math 240! (additions to this web site no longer count towards good deed points)

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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

MondayTuesday is UofT November break

12

Nov 24

HW9

13

Dec 1

Wednesday is a "makeup Monday"! EndofCourse Schedule, Tutorial

F

Dec 8

The Final Exam

Register of Good Deeds

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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 $\implies W_{1}\subset W_{2}\lor W_{2}\subset W_{1}$.
 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 $x+y\in W_{1}\lor x+y\in W_{2}$.
 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 $x\in W_{2}\lor y\in W_{1}$.
 Then $W_{1}\subset W_{2}\lor W_{2}\subset W_{1}$. 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 (2) the last line and (1) the "let" statement to the second last line is not obvious:
(1) Let $x\in W_{1},y\in W_{2}$. [Many lines] Then $x\in W_{2}\lor y\in W_{1}$.
(2) Then $W_{1}\subset W_{2}\lor W_{2}\subset W_{1}$.
Rewrite sentences (1) and (2) into a form that is easier to compare:
(1) $\forall x\in W_{1},\forall y\in W_{2},(x\in W_{2}\lor y\in W_{1})$.
(2) $(\forall x\in W_{1},x\in W_{2})\lor (\forall y\in W_{2},y\in W_{1})$.
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.
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