Download Question 1. a) Prove that √ 10 is irrational. Solution: assume √ 10

Real Analysis 1, Exercise Sheet 1 Solutions, October 9 2012
√
Question 1. a) Prove that 10 is irrational.
√
Solution: assume 10 is
for a contradiction. Then there are
√ rational
a ∈ Z, b ∈ N such that 10 = ab and a, b have no factors in common.
Squaring both sides and rearranging gives 10b2 = a2 . Therefore both 5 and
2 are factors of a2 , and hence of a (using the argument given in lectures).
Using the Fundamental Theorem of Arithmetic, there is an integer n such
that a = 10n. Putting this into 10b2 = a2 gives (on cancelling factors of
10) b2 = 10n2 . Therefore b has a factor of 10 (because b2 does),
so a and b
√
have a factor in common.
/ Q.
√ √This is a contradiction, and so 10 ∈
2 + 5 is irrational. (Hint: Assume by contradiction
b) Hence
√ prove
√ that
p
that 2 + 5 = q , square the equation and then try to simplify first.)
Solution: following the hint, we get
√
p2
2 + 2 10 + 5 = 2 ,
q
from which it follows that
√
1
10 =
2
p2
−7 .
q2
The
√ right-hand side of this equation is a rational number, implying that
√10 ∈√ Q. But this contradicts part (a), from which we conclude that
2 + 5 is irrational.
Question 2. Let x, y ∈ Q and z ∈ R, z ∈
/ Q.
a) Show that x + y ∈ Q.
Solution: write x = a/b and y = p/q. Then x + y = (aq + pb)/bq is
rational because the numerator lies in Z and the denominator in N.
b) Show that x + z ∈
/ Q.
Solution: assume for a contradiction that x + z = r, where r ∈ Q. Then
z = r − x, which lies in Q because Q is closed under subtraction. This
contradicts the hypothesis that z ∈
/ Q. Hence it must be that x + z ∈
/ Q.
c) What about the product xz. Can you be sure it is irrational?
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Solution: yes, it is irrational: if it were rational then there would be r ∈ Q
such that xz = r. But this can be solved for z, giving z = r/x, which is
rational, contrary to our hypothesis that z ∈
/ Q.
Question 3. Let
p
q
be a rational number larger than 43 .
a) Find a rational number between
Solution: take
1
x=
2
and pq .
3
4
3 p
.
+
4 q
b) Find an irrational number between 34 and pq .
Solution: take
1 3
1 p
y = √ + 1− √
24
2 q
Question 4. Given numbers x < y (and you don’t know if they are rational or not), find a rational number between x and y.
Solution: Let d = y − x and note that d > 0 (because y > x). Consider
the real number 2/d: there is a natural number N such that N > 2/d.
(This is the Archimidean property of real numbers. Alternatively, note
that 2/d has a decimal expansion of the form D.E, where D is an integer
and E is a list of natural numbers. Clearly, D is the largest integer smaller
than 2/d, and you can check that D + 1 > D.E. Take N = D + 1.) Since
d > 0, we can rearrange N > 2/d to N d > 2. But N d = N y − N x, so
N y − N x > 2. In other words, N x and N y are real numbers more than 2
units apart. Hence there must be an integer p such that N x < p < N y.
Dividing by N implies that x < p/N < y, as required.
Question 5. Bearing in mind that e is a transcendental number (look up
the definition), show that ln 2 is irrational.
Solution: As is usual when proving irrationality, assume for a contradiction that ln 2 is rational. Thus there exist a ∈ Z and b ∈ N such that
ln 2 = p/q. Hence 2 = ep/q , or 2q = ep . But then e solves the polynomial
equation 2q − xp = 0, which implies that e is algebraic over Z. But e is
transcendental by Hermite’s result, meaning that e cannot satisfy any such
equation. This is a contradiction, and ln 2 ∈
/ Q follows.
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