Math 480
Handout #1
Exercises (page 34)
√
1. Prove the following method for costructing ab. Assume a > b and take a segment AC of length a
and a point B in it such that BC = b. Extend AC to D so that BD = a again. Using compass construct
an isosceles triangle with base AD and sides AE = DE = a. Show that the triangles
EBC and DEB are
√
similar. Hence deduce that if x is the length BE, then xb = xa , which gives x = ab.
2.
theq
two identities
q √
√
p Prove
√
2
2
a ± b = (a+ 2a −b) ± (a− 2a −b) .
√
b
3. The basic approximation in computing square roots is a2 + b ∼ a + 2a
. Prove that
b 2
2
(a ± 2a ) > a ± b.
√
1
1
Use this repeatedly to get 2 ∼ 1 + 13 + 12
− 408
√
3
(Hint: The first application gives 2 > 2. Then 2 = ( 23 )2 − ( 14 ) and so a second application gives
√
1
3
2. A third application gives the required result.)
2 − 12 >
Exercises (page 40)
2. Prove the trigonometric identities used in the Archimedes’ approximation.
1
(a) tan(1 θ ) = sin1 θ + tan
θ,
2
q
(b) sin(1 θ ) = 1 + tan21( θ ) .
2
2
4. Writing a(u) = arctan u, and using the formula
tan A+tan B
tan(A + B) = 1−tan
A tan B
tan A−tan B
(and so tan(A − B) = 1+tan
A tan B )
deduce that
u+v
a(u) + a(v) = a( 1−uv
) and
u+v+w−uvw
a(u) + a(v) + a(w) = a( 1−uv−vw−wu
)
5. Use the previous exercise to verify
(i) arctan 21 + arctan 13 = π4
(ii) arctan 12 + arctan 15 + arctan 18 = π4
1
(iii) 4 arctan 15 − arctan 239
= π4
1
1
1
(iv) 4 arctan 5 − arctan 70 + arctan 99
= π4
37
(Hint: For (iii) and (iv), show that 3 arctan 15 = arctan 55
)
q
p
√
3
7. Show that x =
3 + 5 + 2 is algebraic by constructing an equation of the 12th degree with
integer coefficients satisfied by x.
Exercises (page 51)
2
1. Letq
a, b be positive
q √ integers with a > b. Prove that
√
x = ± a+2 b ± a−2 b
are the roots of the equation
x4 − 2ax2 + b = 0.
2. Verify the following gemetric construction of an irrationalitye like the one in the previous exercise.
Let AB be the diameter of a semicircle of center F and radius 1 and let AB be extended to C so that
BC is the radius. Let CD be the tangent to the semicircle and let E be the midpoint of the arc BD.
Then the length
is given
q CE
q by
p
√
√
√
5 − 2 3 = 5+2 13 − 5−2 13
◦
◦
(Hint: Verify the angle < BCD
deduce using
√ is 30 and hence that the < EF B is also 30 . Hence
2
trigonometry that BE is 2 − 3 and use the theorem of Apollonius in the form CE 2 = 2BE 2 + 1.)
3. Verify the trigonometric identity cos 3θ = 4 cos3 θ − 3 cos θ and hence that x = cos 20◦ satisfies
8x3 − 6x − 1 = 0.
1
The following is a general method for finding the expression for the Fibonacci numbers as er well
as a number of other similar problems. Let (XN ) be a sequence of numbers which are defined by the
recursion relation
XN = pXN −1 + qXN −2 , X0 , X1 arbitrary. It is then clear that once X0 and X1 are given, one can
calculate using this relation the values of X3 , X4 , . . . recursively (hence the name recursion relation).
Here p and q are given number. For the Fiboonacci sequence, p = q = 1 and X0 = 0 and X1 = 1.
5(a) Prove that if we take a trial solution XN = xN , then x must satisfy x2 = px + q
(b) Let a, b be the roots of this quadratic equation. Show that for any constants A and B, the sequence
XN = AaN + BbN satisfies the recursion relation. Choose the constants so that X0 and X1 have given
values.
(c) Deduce from the above the formula for the general term of the Fibonacci√ sequence discussed
in the
√
text, For the Fibonacci sequence with X0 = 1, X1 = 3, show that XN = ( 1+2 5 )N +1 + ( 1−2 5 )N +1
(d) Find a formula for XN where XN = 3XN −1 + XN −2 , X0 = 0, X1 = 1.
Exercises (page 66)
1. Let a, b be roots of the quadratic equation X 2 − AX + B = 0. Find expressions for a2 + b2 and a3 + b3
in terms of A and B. (Hint: Write a2 + b2 as (a + b)2 − 2ab and use the expansion for (a + b)3 to find
an analoguos expression for a3 + b3 )
2. Let Sr = ar + br , where a, b are as in the previous exercise. Prove that S1 Sr−1 = Sr + BSr−2 , S1 = A.
Hence prove by induction on r that each Sr is a polynomial in A and B with integer coeficients. Obtain
by this method the expressions for S5 and S6 in terms of A and B.
4. There are some special equations of degee 4 which can be solved using the solution of quadratic
equations. Prove that
X4 + P X2 + Q = 0
qthe equation
√
1
2
has solutions X = ± −P
2 ± 2 ∆, where ∆ = P − 4Q. Hence show that the solutions of
√ √
5
.
X 4 − 3X 2 + 1 = 0 are ± 6±2
2
Exercises (page 69)
1. For the equation X 3 + 9X = 10. p
show that X =
1 is a root, while ∆ = 52 > 0 and so there del
p
√
√
3
3
Ferro’s formula gives the identity 1 = 2 13 + 5 − 2 13 − 5
2. Find the expression for the root of the cubic equation X 3 + 3X = 6 by del Ferro’s formula.
2