# 14-240/Tutorial-October7

## Boris

#### Subtle Errors in Proofs

Check out these proofs:

##### Proof 1

Let ${\displaystyle W_{1}}$, ${\displaystyle W_{2}}$ be subspaces of a vector space ${\displaystyle V}$.

We show that ${\displaystyle W_{1}\cup W_{2}}$ is a subspace ${\displaystyle \implies W_{1}\subset W_{2}\lor W_{2}\subset W_{1}}$.

Assume that ${\displaystyle W_{1}\cup W_{2}}$ is a subspace.
Let ${\displaystyle x\in W_{1}}$, ${\displaystyle y\in W_{2}}$.
Then ${\displaystyle x,y\in W_{1}\cup W_{2}}$.
Then ${\displaystyle x+y\in W_{1}\cup W_{2}}$.
Then ${\displaystyle x+y\in W_{1}\lor x+y\in W_{2}}$.
Case 1: ${\displaystyle x+y\in W_{1}}$:
Since ${\displaystyle x\in W_{1}}$ and ${\displaystyle W_{1}}$ has additive inverses, then ${\displaystyle (-x)\in W_{1}}$.
Then ${\displaystyle (x+y)+(-x)=y\in W_{1}}$.
Case 2: ${\displaystyle x+y\in W_{2}}$:
Since ${\displaystyle y\in W_{2}}$ and ${\displaystyle W_{2}}$ has additive inverses, then ${\displaystyle (-y)\in W_{2}}$.
Then ${\displaystyle (x+y)+(-y)=x\in W_{2}}$.
Then ${\displaystyle x\in W_{2}\lor y\in W_{1}}$.
Then ${\displaystyle W_{1}\subset W_{2}\lor W_{2}\subset W_{1}}$. Q.E.D.
##### Proof 2

Let ${\displaystyle V=\{(a_{1},a_{2}):a_{1},a_{2}\in R\}}$. Then ${\displaystyle \forall (a_{1},a_{2}),(b_{1},b_{2})\in V,\forall c\in R}$, define

${\displaystyle (a_{1},a_{2})+(b_{1},b_{2})=(a_{1}+2b_{1},a_{2}+3b_{2})}$ and ${\displaystyle c(a_{1},a_{2})=(ca_{1},ca_{2})}$.

We show that ${\displaystyle V}$ is not a vector space over ${\displaystyle R}$.

We show that ${\displaystyle V}$ is not commutative.
Let ${\displaystyle (a_{1},a_{2})=(0,0)}$.
Then ${\displaystyle (0,0)+(b_{1},b_{2})=(2b_{1},3b_{2})\neq (b_{1},b_{2})}$.
Then ${\displaystyle V}$ is not commutative.
Then ${\displaystyle V}$ 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 ${\displaystyle x\in W_{1},y\in W_{2}}$. [Many lines] Then ${\displaystyle x\in W_{2}\lor y\in W_{1}}$.

(2) Then ${\displaystyle W_{1}\subset W_{2}\lor W_{2}\subset W_{1}}$.

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

(1) ${\displaystyle \forall x\in W_{1},\forall y\in W_{2},(x\in W_{2}\lor y\in W_{1})}$.

(2) ${\displaystyle (\forall x\in W_{1},x\in W_{2})\lor (\forall y\in W_{2},y\in W_{1})}$.

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