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MATH 245: QUIZ 4 SOLUTIONS
1)
Let
c
be the conjecture "
is irrational".
Let's use a proof by contradiction to prove
c
.
Assume
¬
c
, the conjecture "
is rational", is true.
Then, there exist relatively prime
a
∈
Z
and
b
∈
Z
b
≠
0
( )
such that
=
a
b
. (The "relatively prime" condition ensures that the fraction is
in lowest terms.)
3
=
a
b
⇒
3
=
a
2
b
2
⇒
3
b
2
=
a
2
b
2
∈
Z
, so 3 divides
a
2
and is, therefore, a prime factor
of
a
2
. Any prime factor of
a
2
must also be a prime factor
of
a
, so 3 must also divide
a
. That means that
∃
c
∈
Z
:
a
=
3
c
.
⇒
3
b
2
=
3
c
( )
2
⇒
3
b
2
=
9
c
2
⇒
b
2
=
3
c
2
c
2
∈
Z
, so 3 divides
b
2
and is, therefore, a prime factor
of
b
2
. Any prime factor of
b
2
must also be a prime factor
of
b
, so 3 must also divide
b
.
We've shown that 3 must divide both
a
and
b
, which
contradicts the relative primality of
a
and
b
.
Therefore,
¬
c
is false, and
c
is true.
Another proof:
If
were rational, then it would be a rational root of the polynomial
x
2
−
3
. However, the Rational Roots Theorem suggests that the only
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 Spring '11
 staff
 Math

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