P2 Chapter 9 Introduced in P2 are the specific methods of “proof by contradiction” and “disproof by counter-example”. Note that the use of these methods may be required in either the P2 or the P3 examination. An asterisk (*) next to the question number suggests that the question in this form might not be appropriate for an external examination. (See the end of the individual solution for further explanation in each case.)

P2 EXERCISE 9A 1. Counter-example: Take x = 56 π : sin 56 π = 12

2. Counter-example: Take u = −1, v = −2, x = 3, y = 1

u > v, and x > y, but ux = −3 and vy = −2, so ux < vy

3. Counter-example: Take x = −5 : x < 4, but x2 = 25 > 16

4. Counter-example: Take a = −1 : “b2 − 4ac” = 1 − 4 = −3

“b2 − 4ac” < 0, so the roots are not real.

5. Counter-example: Take x = 4 : x2 + 5x = 16 + 20 = 36

36 > −6, so the set of values −2 < x <3 is incomplete or incorrect.

6. Counter-example: Take n = 41 : f (41) = 412 + 41 + 41 = 41 × 43

So f (41) is not prime

7. Counter-example: Take a = 1, b = −4, c = 4 :

Repeated real root 2

(Roots are real (and equal) whenever b2 = 4ac)

8. All interior angles equal does not imply all sides equal,

e.g. rectangle.

All sides equal does not imply all interior angles equal,

e.g. rhombus.

9. Counter-example: Take n = 4 : f (n) = 5 × 6 × 7 = 210

210 is not divisible by 12

10. Counter-examples are found in the second quadrant,

e.g. Take x = 120° and y = 150° : sin x > sin y, but x < y

11. Suppose that for all x > 1 , xx

+ ≤1 2

Multiplying by x : x2 + 1 ≤ 2x (since x is positive)

x2 − 2x + 1 ≤ 0 ⇒ (x − 1)2 ≤ 0

A contradiction. (Since x > 1 , even (x − 1)2 = 0 is not possible)

So, for all x > 1, xx

+1 > 2 .

12.* x2 − 2x = x (x − 2)

If x ≤ 0, x (x − 2) = (negative or zero) × (negative) ≥ 0

x2 − 2x ≥ 0, x2 ≥ 2x Contradiction

If x ≥ 2, x (x − 2) = (positive) × (positive or zero) ≥ 0

x2 − 2x ≥ 0, x2 ≥ 2x Contradiction

But if 0 < x < 2, x (x − 2) = (positive) × (negative) < 0

(*) Other methods of proof for this question would be equally valid in

a P2 examination.

13. Suppose there exist integers p and q such that pq

2

22= .

Then pq

⎛⎝⎜

⎞⎠⎟

2

= 2 , so pq= 2

A contradiction, since √2 is irrational.

So there exist no integers p and q such that pq

2

22= .

14. Suppose the number of rational numbers between 0 and 1 is finite.

Then there exists a largest such rational number a, (a < 1).

But consider b = a a+

−( )12

(half way between a and 1):

If a = pq

, where p and q are integers,

b = a a+

−( )12

= a + 12

=

pq p q

q

+=

+1

2 2

Both (p + q) and 2q are integers, so b is rational and b > a

So a is not the largest such rational number. Contradiction.

So the number of rational numbers between 0 and 1 is infinite.

15. Suppose that √3 is rational.

Let √3 = ab , where a and b are integers having no common factor.

ab

2

23= ⇒ a2 = 3b2 ⇒ a2 is a multiple of 3

⇒ a is a multiple of 3, say a = 3c (integer c)

(3c)2 = 3b2 ⇒ 9c2 = 3b2 ⇒ b2 = 3c2

So b2 is a multiple of 3, which implies that b is a multiple of 3.

So a and b have a common factor 3. Contradiction.

So √3 is irrational.

16. x3 − 3x2 − 2x + 3 = 0 has a root in the interval (N, N + 1)

Consider f (x) = x3 − 3x2 − 2x + 3

f (−2) = −13 and f (−1) = 1

f (0) = 3 and f (1) = −1

f (3) = −3 and f (4) = 11

For each pair of consecutive integer values of x considered above,

the sign change in f (x) indicates a root in the interval.

So there are 3 possible values of N (−2, 0 and 3).

(A cubic equation has at most 3 roots).

17.* ( )2 11

rr

n−

=∑ = 1 + 3 + 5 + 7 +… + (2n − 1)

This is an arithmetic series with n terms, a = 1 and d = 2.

Sum = 12 n (2 + 2 (n − 1)) = n2

(*) It would be reasonable here to assume the sum formula for an

arithmetic series, unless told otherwise.

18. x2 + 4x + 5 > 0 (x + 2)2 − 4 + 5 > 0

(x + 2)2 + 1 > 0

Since (x + 2)2 ≥ 0 for all real values of x,

(x + 2)2 + 1 > 0 for all real values of x.

19.* The geometric series has a = 13 and r = −19

Sum to infinity = ar1−

= 13 ÷ 109 = 3

10 .

(*) As in question 17, the standard formula would be used unless the

question demanded otherwise.

20.* Sum to n terms = nn 2 1+

(a) S1 = 12 , so the first term of the series is 12

( )( ) ( )( )

( ) ( )( )T S S n

nn

n

n n n n n

n nn n n= − =

+− −

− +=

− + − − +

+ − +−1 2 2

2 2

2 211

1 1

2 2 1 1

1 1 1

( )

( ) ( )( ) [ ] ( )[ ]T

n n

n n

n

n nn = −

− −

+ − += −

−⎛⎝⎜

⎞⎠⎟

−⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ − +

2

2 2

2

2 2

1

1 1 1

12

54

1 1 1

If n ≥ 2, (n − 12 )2 ≥ 94 , so each of the square brackets above

has a positive value, so the value of the expression is negative.

So all terms of the series after the first are negative.

(a) An alternative argument:

S2 = 25 , so T2 = 25 − 12 = − 110

The sum to n terms is nn n

n

2 11

1+=

+⎛⎝⎜

⎞⎠⎟

Considering positive integer values of n, it can be seen that as n

increases, ( )nn

+1 increases, so 1

1nn

+⎛⎝⎜

⎞⎠⎟

decreases.

So, since T2 is negative and the sum is decreasing, all subsequent

terms must also be negative.

(b) ( )

T S Sn

nn

nn n n= − =

+−

−

− +−1 2 21

11 1

= 11nn

+⎛⎝⎜

⎞⎠⎟

− ( ) ( )

1

1 11

nn

− +−

As n → ∞, 1n

→ 0, so 11nn

+⎛⎝⎜

⎞⎠⎟

→ 1n

→ 0

and similarly 11n −

→ 0, so ( ) ( )

1

1 11

nn

− +−

→ 11n −

→ 0

So, as n → ∞, Tn → 0

(*) This question is too demanding for a P2 examination.