1. (a) Find the (complex) roots of the quadratic equation x2−5x+7−i = 0.
(b) Find the roots of the quartic equation x4+x2+1 = 0.
(c) Find the roots of the equation 2x4−4x3+3x2+2x−2 = 0, given that one of them is 1+i.
2. Use the method given in this chapter for solving cubics to find the roots of the equation x3− 6x2+13x−12 = 0.
Now notice that 3 is one of the roots. Reconcile this with the roots you have found.
3. Solve x3− 15x − 4 = 0 using the method for solving cubics.
Now cleverly spot an integer root. Deduce that cos 1
4. Factorize x5+1 as a product of real linear and quadratic polynomials.
5. Show that cos2π9 is a root of the cubic equation 8x3− 6x + 1 = 0.
Find the other two roots and deduce that cos 2π
6. (a) Factorize x2n+1− 1 as a product ofreal linear and quadratic polyno-mials.
(b) Write x2n+x2n−1+∙∙∙ + x + 1 as a product of real quadratic polyno-mials.
(c) Letω=e2πi/2n+1. Show that ∑ωi+ j=0, where the sum is over all i and j from 1 to 2n+1 such that i < j.
POLYNOMIAL EQUATIONS 57 7. (a) Find the value of k such that the roots of x3+6x2+kx − 10 = 0 are in arithmetical progression (i.e., areα,α+d,α+2d for some d). Solve the equation for this value of k.
(b) If the roots of the equation x3− x − 1 = 0 areα,β,γ, find a cubic equation having rootsα2,β2,γ2and also a cubic equation having roots
α1,β1,1γ.
(c) Given that the sum of two of the roots of the equation x3+px2+ p2x+r = 0 is 1, prove that r = (p+1)(p2+p+1).
(d) Solve the equation x4− 3x3− 5x2+17x − 6 = 0, given that the sum of two of the roots is 5.
8. Critic Ivor Smallbrain and his friend Polly Gnomialle have entered a competition to try to qualify to join the UK team for the annual Film Critics Mathematical Olympiad. The qualification competition consists of one question. Here it is:
Find all real or complex solutions of the simultaneous equations x+y+z = 3
x2+y2+z2=3 x3+y3+z3=3
The first pair to correctly answer the question gets into the team. Ivor overhears his arch-rival Greta Picture, who has also entered the compe-tition, whispering to her partner, “Let’s look for a cubic equation which has roots x,y and z.”
Can you help Ivor and Polly win?
Chapter 8
Induction
Consider the following three statements, each involving a general positive in-teger n:
(1) The sum of the first n odd numbers is equal to n2. (2) If p > −1 then (1+ p)n≥ 1 + np.
(3) The sum of the internal angles in an n-sided polygon is (n−2)π. [A polygon is a closed figure with straight edges, such as a triangle (3 sides), a quadrilateral (4 sides), a pentagon (5 sides), etc.]
We can check that these statements are true for various specific values of n. For instance, (1) is true for n = 2 as 1 + 3 = 4 = 22, and for n = 3 as 1 + 3 + 5 = 9 = 32; statement (2) is true for n = 1 as 1 + p ≥ 1 + p, and for n = 2 as (1+ p)2=1+2p+ p2≥ 1 + 2p; and (3) is truefor n = 3 as the sum of the angles in a triangle isπ, and for n = 4 as the sum of the angles in a quadrilateral is 2π.
But how do we go about trying to prove the truth of these statements for all values of n?
The answer is that we use the following basic principle. In it we denote by P(n) a statement involving a positive integer n; for example, P(n) could be any of statements (1), (2) or (3) above.
Principle of Mathematical Induction
Suppose that for each positive integer n we have a statement P(n). If we prove the following two things:
(a) P(1) is true;
(b) for all n, if P(n) is true then P(n+1) is also true;
then P(n) is true for all positive integers n.
The logic behind this principle is clear: by (a), the first statement P(1) is true. By (b) with n = 1, we know that P(1) ⇒ P(2), hence P(2) is true. By (b) with n = 2, P(2) ⇒ P(3), hence P(3) is true; and so on.
59
60 A CONCISE INTRODUCTION TO PURE MATHEMATICS The principle may look a little strange at first sight, but a few examples should clarify matters.
Example 8.1
Let us try to prove statement (1) above using the Principle of Mathe-matical Induction. HereP(n) is the statement that the sum of the first n odd numbers is n2. In other words:
P(n) : 1+3+5+∙∙∙+2n−1 = n2. We need to carry out parts (a) and (b) of the principle.
(a)P(1) is true, since 1 = 12. (b) SupposeP(n) is true. Then
1+3+5+∙∙∙+2n−1 = n2. Adding2n+1 to both sides gives
1+3+5+∙∙∙+2n−1+2n+1 = n2+2n+1 = (n+1)2,
which is statementP(n+1). Thus, we have shown that P(n) ⇒ P(n+1).
We have now established parts (a) and (b). Hence by the Principle of Mathematical Induction,P(n) is true for all positive integers n.
The phrase “Principle of Mathematical Induction” is quite a mouthful, and we usually use just the single word “induction” instead.
Example 8.2
Now let us prove statement (2) above by induction. Here, for n a positive integerP(n) is the statement
P(n) : if p > −1 then (1+ p)n≥ 1 + np . For (a), observeP(1) is true, as 1+ p ≥ 1+ p.
For (b), supposeP(n) is true, so (1 + p)n≥ 1 + np. Since p > −1, we know that1+ p > 0, so we can multiply both sides of the inequality by 1+ p (see Example 4.3) to obtain
(1+ p)n+1≥ (1 + np)(1 + p) = 1 + (n + 1)p + np2.
Sincenp2≥ 0, this implies that (1 + p)n+1≥ 1 + (n + 1)p, which is state-mentP(n+1). Thus we have shown P(n) ⇒ P(n+1).
Therefore, by induction,P(n) is true for all positive integers n.
Next we attempt to prove the statement (3) concerning n-sided polygons.
There is a slight problem here. If we naturally enough let P(n) be statement (3),
INDUCTION 61 then P(n) makes sense only if n ≥ 3; P(1) and P(2) make no sense, as there is no such thing as a 1-sided or 2-sided polygon. To take care of such a situation, we need a slightly modified Principle of Mathematical Induction:
Principle of Mathematical Induction II
Let k be an integer. Suppose that for each integer n ≥ k we have a statement P(n). If we prove the following two things:
(a) P(k) is true;
(b) for all n ≥ k, if P(n) is true then P(n+1) is also true;
then P(n) is true for all integers n ≥ k.
The logic behind this is the same as explained before.
Example 8.3
Now we prove statement (3). Here we havek = 3 in the above principle, and forn ≥ 3, P(n) is the statement
P(n) : the sum of the internal angles in an n-sided polygon is (n−2)π. For (a), observe thatP(3) is true, since the sum of the angles in a triangle isπ= (3−2)π.
Now for (b). SupposeP(n) is true. Consider an (n+1)-sided polygon with cornersA1,A2, . . . ,An+1:
Draw the lineA1A3. ThenA1A3A4. . .An+1is ann-sided polygon. Since we are assumingP(n) is true, the internal angles in this n-sided polygon add up to (n−2)π. From the picture we see that the sum of the angles in the (n+1)-sided polygon A1A2. . .An+1is equal to the sum of those in A1A3A4. . .An+1 plus the sum of those in the triangleA1A2A3, hence is
(n−2)π+π= ((n+1)−2)π .
We have now shown thatP(n) ⇒ P(n+1). Hence, by induction, P(n) is true for alln ≥ 3.
62 A CONCISE INTRODUCTION TO PURE MATHEMATICS The next example also uses the slightly modified Principle of Mathematical Induction II. In it, for a positive integer n we define
n! = n(n−1)(n−2)...3∙2∙1 ,
the product of all the integers between 1 and n. The symbol n! is usually referred to as n factorial. By convention, we also define 0! = 1.
Example 8.4
For which positive integersn is 2n<n! ?
Answer Let P(n) be the statement that 2n<n!. Observe that 21>1!, 22>2!, 23>3!, 24<4!, 25<5! ,
so P(1),P(2),P(3) are false, while P(4),P(5) are true. Therefore, it seems sensible to try to prove P(n) is true for all n ≥ 4.
First, P(4) is true, as observed above.
Now suppose n is an integer with n ≥ 4, and P(n) is true. Thus 2n<n!
Multiplying both sides by 2, we get
2n+1<2(n!) .
Since 2 < n + 1, we have 2(n!) < (n + 1)n! = (n + 1)! and hence 2n+1<
(n + 1)!. This shows that P(n) ⇒ P(n + 1). Therefore, by induction, P(n) is true for all n ≥ 4.