PRELIMINARY ALGEBRA
1.11 Show that the equation is equivalent to sin(5θ/2) sin(θ)sin(θ/2) = 0.
Solutions are− 4 π/ 5 ,− 2 π/ 5 , 0 , 2 π/ 5 , 4 π/ 5 ,π.Thesolutionθ= 0 has multiplicity
3.
1.13 (a) A circle of radius 5 centred on (− 3 ,−4).
(b) A hyperbola with ‘centre’ (3,−2) and ‘semi-axes’ 2 and 3.
(c) The expression factorises into two lines,x+2y−3=0and2x+y+2=0.
(d) Write the expression as (x+y)^2 =8(x−y) to see that it represents a parabola
passing through the origin, with the linex+y= 0 as its axis of symmetry.
1.15 (a)
5
7(x−2)+
9
7(x+5), (b)−4
3 x+
4
3(x−3).
1.17 (a)
x+2
x^2 +4−
1
x− 1, (b)x+1
x^2 +9+
2
x^2 +1.
1.19 (a) 10, (b) not defined, (c)−35, (d)−21.
1.21 Look for factors common to then=Nsum and the additionaln=N+1 term,
so as to reduce the sum forn=N+1 to a single term.
1.23 Write 3^2 nas 8m−7.
1.25 Use the half-angle formulae of equations (1.32) to (1.34) to relate functions of
θ/ 2 kto those ofθ/ 2 k+1.
1.27 Divisible fork=1, 2 ,...,p− 1 .Expand (n+1)pasnp+
∑p− 1
1pCknk+ 1. Apply
the stated result forp= 5. Note thatn^5 −n=n(n−1)(n+1)(n^2 + 1); the product
of any three consecutive integers must divide by both 2 and 3.
1.29 By assumingx=p/qwithq= 1, show that a fraction−pn/qisequaltoan
integeran− 1 pn−^1 +···+a 1 pqn−^2 +a 0 qn−^1. This is a contradiction, and is only
resolved ifq= 1 and the root is an integer.
(a) The only possible candidates are± 1 ,± 2 ,±4. None is a root.
(b) The only possible candidates are± 1 ,± 2 ,± 3 ,±6. Only−3isaroot.
1.31 f(x) can be written asx(x+1)(x+2)+x(x+1)(x−1). Each term consists of
the product of three consecutive integers, of which one must therefore divide by
2 and (a different) one by 3. Thus each term separately divides by 6, and so
therefore doesf(x). Note that ifxis the root of 2x^3 +3x^2 +x−24 = 0 that lies
near the non-integer valuex=1.826, thenx(x+ 1)(2x+ 1) = 24 and therefore
divides by 6.
1.33 Note that, e.g., the condition for 6a 4 +a 3 to be divisible by 4 is the same as the
condition for 2a 4 +a 3 to be divisible by 4.
For the necessary (only if) part of the proof setn=1, 2 ,3 and take integer
combinations of the resulting equations.
For the sufficient (if) part of the proof use the stated conditions to prove the
proposition by induction. Note thatn^3 −nis divisible by 6 and thatn^2 +nis
even.