MATH 2400: PRACTICE PROBLEMS FOR EXAM 1
PETE L. CLARK
1) Find all real numbers
x
such that
x
3
=
x
. Prove your answer!
Solution:
If
x
3
=
x
, then 0 =
x
3
−
x
=
x
(
x
+ 1)(
x
−
1).
Earlier we showed
using the field axioms that if if a product of numbers is equal to zero, then at least
one of the factors must be zero. So we have
x
= 0,
x
+ 1 = 0 or
x
−
1 = 0, i.e.,
x
= 0
,
±
1. Conversely, these three numbers do indeed satisfy
x
3
=
x
.
2) a) Prove that
√
6 is an irrational number.
You may use the fact that if an
integer
x
2
is divisible by 6, then also
x
is divisible by 6. (For “extra credit”, prove
the fact of the previous sentence using the uniqueness of prime factorizations.)
Solution: let
N
be a positive integer. We claim that if
N
2
is divisible by 6, then
so is
N
. Indeed by the uniqueness of prime factorization, since
N
2
is divisible by 2
and 3, 2 and 3 must each appear in the prime factorization of
N
.
Now we show that
√
6 is irrational. Seeking a contradiction, we suppose there
are integers
a, b
,
b
̸
= 0, such that
√
6 =
a
b
.
We may assume that
a
and
b
have
no common prime divisor.
Squaring both sides and clearing denominators gives
6
b
2
=
a
2
, so
a
2
is divisible by 6. By the above paragraph this means
a
is divisible
by 6: say
a
= 6
A
for some integer
A
. Thus 6
b
2
=
a
2
= (6
A
)
2
= 36
A
2
, so 6
A
2
=
b
2
.
Thus
b
2
is divisible by 6 and
b
is divisible by 6, contradicting the fact that
a
and
b
are relatively prime.
Comment:
It is not in fact necessary to establish the fact that
N
2
divisible by
6 implies
N
divisible by 6.
Indeed, we can use the corresponding fact for divisi
bility by 2: if 6
b
2
=
a
2
then
a
2
is even, so
a
is even, so
a
= 2
A
for some
A
∈
Z
.
Thus
6
b
2
=
a
2
= (2
A
)
2
= 4
A
2
,
and so
3
b
2
= 2
A
2
.
So 3
b
2
is even. Since 3 is odd and the product of two odd numbers is odd,
b
2
must
be even and thus
b
is even. So
a
and
b
are both even, contradicting our assumption
that they are relatively prime.
b) Show that if
x
2
is an irrational number, so is
x
.
First Solution: Seeking a contradiction we suppose that
x
is rational, so
x
=
a
b
.
Then
x
2
=
a
b
a
b
=
a
2
b
2
would be rational, contradiction.
Second Solution: Rather than presenting the above argument as a proof by contra
diction, we may also phrase it as a proof by
contrapositive
. Namely, the statement
1
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2
PETE L. CLARK
A
=
⇒
B
is logically equivalent to its contrapositive statement not
A
=
⇒
not
B
.
In this case, the contrapositive of the statement to be proved is: if
x
is rational, so
is
x
2
. Notice that this is exactly what we showed above.
c) Show that
√
2 +
√
3 is irrational.
Solution: Let
x
=
√
2+
√
3. By part b), it suﬃces to show that
x
2
is irrational. Now
x
2
= (
√
2+
√
3)
2
= 2+2
√
6+3 = 5+2
√
6. If 5+2
√
6 =
a
b
, then 2
√
6 =
a
b
−
5 =
a
−
5
b
b
and thus
√
6 =
a
−
5
b
2
b
would be a rational number, contradicting part a).
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 Fall '11
 Clark
 Calculus, Real Numbers, pete l. clark

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