14-240/Tutorial-October7

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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, y \in W_2 }[/math].

Then [math]\displaystyle{ x, y \in W_1 \cup W_2 }[/math] and [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) = (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 equivalence of these two statements.

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

We can rewrite (1) and (2) to see their difference from each other:

(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_1, y \in W_2) }[/math].