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 $W_{1}$, $W_{2}$ 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 $x,y\in W_{1}\cup W_{2}$ and $x+y\in W_{1}\cup W_{2}$.

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?