Math 25: Solutions To Homework # 3 Worksheet - Dartmouth College
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4Math 25: Solutions to Homework # 3
(3.5 # 44) Show that
5 is irrational.
(a) Suppose
5 is rational. Then we can write
5 = a/b where (a, b) = 1 and b = 0.
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3
3
3
3
Then 5 = a
/b
, so 5b
= a
. Now 5 a
, so 5 a. Then we can write a = 5k for some integer
3
3
3
k, so 5b
= 125k
, and hence 5
b
, so 5
b. But this is a contradiction since (a, b) = 1.
Therefore
5 is irrational.
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(b) Since
5 is not an integer, and it is the root of the polynomial x
5, it is irrational,
by Theorem 3.18.
(3.5 # 74) Show that if p is prime and 1
k < p, then the binomial coefficient
is divisible
by p.
The binomial coefficient
p
p!
1 2
p
=
=
.
k
k!(p
k)!
1 2
k 1 2
(p
k)
Since k < p, all the factors in the denominator are less than p, so they do not cancel the p
in the numerator. Therefore, p divides
.
(3.6 # 16) Show that if a is a positive integer and a + 1 is an odd prime, then m = 2 for
some positive integer n.
Suppose that a + 1 is an odd prime. If m = k with > 1 odd, then we can factor
(
1)
(
2)
a + 1 = (a + 1)(a
a
+
a + 1).
Since k < m, a +1 < a +1, and since a > 0, a +1 > 1, so this is a nontrivial factorization,
and hence a contradiction. Therefore m must have no odd factors, so it must be of the form
m = 2 .
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(3.6 # 18) Use the fact that every prime divisor of F
= 2
+1 is of the form 2
k+1 = 64k+1
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to verify that F
is prime.
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Any prime factor of F
must be of the form 64k + 1, and must be less than or equal to
4
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[ 65, 537] = 256 = 2
. Then 64 + 1 = 65 is not prime, 64 2 + 1 = 129 is not prime, and
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64 3 + 1 = 193 F
. The next possible factor 64 4 + 1 = 2
+ 1 is too big, so F
is prime.
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