Every geometric figures with four right angles is a square. (5)
A
rectangle has four right angles and isnt a square
For every integer n, the number 3(n2 + 2n+ 3) 2n2 is a perfect square. (5)
3(n2
+ 2n +3) 2n2 =
3n2 + 6n + 9 2n2 =
n2 + 6n + 9 =
(n+3) (n+3) =
(n+3) 2
for n an even integer, 2n - 1 is not prime.
for n = 3, 2n = 8.
2n 1 = 7, 7 is prime.
statement disproved
by finding a counter example
The difference of the cubes of two consecutive integers is odd.
(2n+1)(2n+1)(2n+1)-
(2n)(2n)(2n)
(4n2+4n+1)(2n+1) 8n3
8n3+ 8n2 +
2n + 4n2 + 4n + 1 8n3
12n2 + 6n +
1
2(6n2 + 3n)
+ 1
2k + 1 which is odd
5+10+15+...+5n = [(5n)(5n+1)]/2
n = 1, 5*1 = 5 5(1)(1+1) = 5*2 =
5 proved
2 2
for k, 5+10 + 15 + ... + 5k = [5k(k+1)] /2 assumed true
k+1 5 + 10 + 15 + ... + 5k + 5(k+1) = [5(k+1) (k+1+1 )]/2 to be proved
5 + 10 + 15
+ ... + 5k +
5(k+1) =
[5k(k+1) ]/2 + 5(k+1) =
[5k(k+1) + 2*5*(k+1)]/ 2
5(k+1) [ (k+2)]
2
n = 3 n2= 9 >= 2(n) + 3 = 9
assume for 3 <= r <=
k k2 >= 2(k) + 3
prove (k+1)2 >=
2(k+1) + 3
(k+1)2 = k2 + 2k + 1 but k2 > 2k+3
>= 2k + 3 + 2k + 1 = 4k + 4 LHS
RHS = 2k +
2 + 3 or 2k + 5
subtracting 2k from
each side
LHS = 2k +
4 RHS = 5
since k
>= 3 2k + 4>= 10 and
10 >= 5 QED
7n-2n is divisible by 5
n = 1 7n-2n = 71-21 = 7 - 2 = 5
assume 7k-2k = 5m
prove 7k+1-2k+1 = 5x
7k+1
= 7(7k) so 7k+1-2k+1
= 7(7k) - -2k+1
but 7k = 5m
- 2k so 7(7k) -2k+1 =
7(5m - 2k) -2k+1
== 7(5m) -
7(2k) - 2(2k) =
7(5m) - 2k(7 - 2) = 5(7m) - 2k(5)
1. 2 belongs to T
if X belongs to T, so does X + 3 and 2*X.
Circle the following that belong to T.
a. 6 b. 7 c. 19 d. 12
L(1) = 1
L(2) = 3
L(n) = L(n-1) + L (n-2) for n ≥ 3
b. Write the next three terms of the sequence (3)
L(3) =
4
L(4) = 7
c. Prove that L(n) = F(n + 1) + F (n -1) for n ≥ 3
for lower bound
n = 3
L(3) = F(4) + F(2) = L(3) = 4
F(4) + F(2) = 3 + 1 = 4
assume 3 <= r <= k
L(k) = F(k+1) + F(k-1)
prove: L(k+1) =
F(k+2) + F(k)
by definition L(k) = L(k-1) + L(k-2)
L(k+1) = L(k+1-1) + L(k+1-2)
L(k+1) = L(k) + L(k-1)
L(k) = F(k+ 1) + F(k-1) L(k-1) = F(k-1 + 1) + F(k-1 -1) = F(k) + F(k-2)
L(k+1) = L(k) + L(k-1)
= F(k+1) + F(k-1) + F(k) + F(k-2)
= F(k+1) + F(k) +
F(k-1) + F(k-2)
= F(k+2) + F(k)
F(k+2) + F(k) =
F(k+2) + F(k) q.e.d
½
b. What is the probability of exactly three (3) girls? (5)
bbbb bbbg
bbgb bbgg
bgbb bgbg
bggb bggg 4/16
gbbb gbbg
gbgb gbgg
ggbb ggbg gggb gggg
a. The third term in (a + b) 7 (5)
21a5b2
b. The last term in (ab + 3x) 6 (5)
(3x6)
P(A and B) = 8%
P(B)
34%
P (A or B)
= P(A) + P(B)
P(A and B) = 17 + 34 8 = 43%
100% - 43%
= 57%
Section 4.1 # 8b (5)
r = { (0,1), (1,0), (2,4), (4,2), (4,6), (6,4)}
x r y « x is the brother of y
where T = {1,3,5,7,9} and
where f = { (6,3), (8,1), (0,3), (4,5), (2,7)}
Task |
Prerequisite Tasks |
Time to Perform |
A |
E |
3 |
B |
C,D |
5 |
C |
A |
2 |
D |
A |
6 |
E |
None |
2 |
F |
A,G |
4 |
G |
E |
4 |
H |
B,F |
1 |
17
E A D B H
E A C D G F B H or E A D G F C B H or
...