Hints for Proofs
1. First, write out the hypothesis and conclusion in mathematical terms (label variables, write
out equations, etc.).
Example: Consider the following Theorem: The intersection of two subspaces of a vector
space is also a subspace.
Start out by labeling the big vector space by Vand the two subspaces by Yand Z.
Hypothesis:Yand Zare subspaces of a vector space V.
Conclusion:Y∩Zis also a subspace of V.
2. Assume only the hypothesis and nothing else (except previous theorems). Don’t mix up
different types of objects, like sets and elements.
3. Proceed logically at each step (make sure each step follows from the previous step).
Example: The implication y2=x2⇒y=xis not correct for all x, y ∈R.
4. Use previous theorems and results.
5. It may be necessary to divide the proof into several cases.
Example: Consider the following Theorem:Ifad −bc 6= 0, then the system ax +by =0
cx +dy =0
has exactly one solution x=y=0.
Proof : (sketch) First suppose a=0. Thenband care nonzero, which implies that y=0
and then x= 0. Now suppose that a6=0. Thenx=−by
afrom the first equation, and the
proof can be completed by substituting this value into the second equation.
6. Maybe use proof by contradiction. If you are trying to prove “P⇒Q”, there are two ways
to do this: (a) prove the contrapositive “not Q⇒not P”, or (b) assume that “P⇒Q” is false,
i.e., that Pis true and Qis false, and proceed until you contradict something, such as a previous
theorem or some obvious fact like 0 6= 1. The following example illustrates (a).
Example: Consider the following Theorem:Ifdim(V)=n,thenanysetofnvectors which
spans Vmust also be linearly independent.
Proof :LetS={s
1
,...,s
n}be a set of nvectors which spans V, and suppose that Sis
linearly dependent. Then by an earlier result, we can remove some vector sifrom Ssuch
that the resulting set S1still spans V. But the definition of dimension then implies that
dim(V)≤n−1<n, a contradiction to the hypothesis. Therefore, Smust be linearly
independent.
Here is a simple example which illustrates (b).
Example: Consider the following Theorem:Iffis the function defined by f(x)=e
xfor all
x∈R,thenf(x)6=0 for allx∈R.
Proof : Suppose f(x0)=0forsomex
0∈R.Then0=f(x
0
)f(−x
0
)=e
x
0
e
−x
0=e
x
0
−x
0=
e
0= 1, a contradiction. Therefore, f(x)6=0forallx∈R.
7. Look at simple cases to get an idea of why the implication might be true or false. Then look
at the general case.
Example: To show th a t d i m (M(m, n)) = mn,lookatM(2,3) first.
8. Be creative. Look at the problem in different ways. Maybe write down equivalent statements.
Example: The statement “the system ax +by =c
dx +ey =fhas a unique solution” is equivalent to
“the lines ax +by =cand dx +ey =fintersect in a single point”.
9. Don’t give up too easily.
