Theorem
The
square root of any
prime number is
irrational.
Proof
Let
p
be
prime.
Suppose that
p √
is
rational.
Then there exist
natural numbers m
and
n
such that:
|
|
|
| p √
| =
|
| m n
|
|
| | |
|
| ⟹
|
| p
| =
|
| m 2 n 2
|
|
| | |
|
| ⟹
|
| n 2 p
| =
|
| m 2
|
|
| | |
Any
prime in the
prime factorizations of
n 2
and
m 2
must occur an
even number of times because they are
squares.
Thus,
p
must occur in the prime factorization of
n 2 p
an
odd number of times.
Therefore,
p
occurs as a factor of
m 2
an odd number of times, a
contradiction.
So
p √
must be
irrational.
ProofWiki
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