Math 111 Test Template Page 2
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1
2(b) Is R symmetric?
Yes, because if (a, b) ∈ R, then a + b < 5, then b + a < 5, so (b, a) ∈ R.
(c) Is R transitive?
No, because e.g. (2, 1) ∈ R and (1, 3) ∈ R, but (2, 3) ∈ R since 2 + 1 < 5, 1 + 3 < 5,
but 2 + 3 < 5.
7. Prove or disprove.
√
√
The number
3 +
5 is irrational.
√
√
The statement is true. We will prove it by contradiction. Suppose x =
3 +
5 is
m
rational. Then x =
for some m, n ∈ Z, n = 0. Since x > 0, m = 0. We have
n
√
√
√
√
√
2
2
2
3 =
5. Then (x
3)
= (
5)
, so x
2x
3 + 3 = 5. This implies that
x
2
2
2
√
2n
2
2
2
2
2
m
m
2n
x
m
2
2
2
2
3 =
=
=
=
. Since m
2n
, 2mn ∈ Z and 2mn = 0,
n
n
2m
2x
2
2mn
m
√
√
n
n
3 is rational. However, by a homework problem
3 is irrational. Contradiction.
Therefore x is irrational.
2
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