Download Lesson 1 - Suffolk Maths

Proof Assignment Solutions
1.
2.
3.
4.
5.
6.
7.
8.
The Need For Proof
Inductive Reasoning
Deductive Reasoning
Conditional Statements and Proofs By Counterexample
Integer Property Proofs
Deductive Geometric Proofs and Indirect Proofs
Coordinate Geometry Proofs
Proofs By Mathematical Induction
Lesson 1
The Need For Proof
1. To average 100 km/h, the time taken to the 500 km round trip would be 500km/ 100km/h or
5 hours. However the time taken to gert to Saskatoon is 250km/ 50km/h or 5 hours.
Therefore the motorist would have to return in 0 hours. That's impossible. There is no speed
at which the motorist could return for this to happen.
2. C = d
therefore C/ = d
when C = 40,000km, d = 40,000km / = 12732.39545km or 12732395.45m
when C = 40000.002km, d=40,000.002 / = 12732.39608km or 12732296.08m
The difference in diameters is .633971m or 63.4cm. If the ring remains the same distance
from the earth there would be 63.4cm/2 or 31.7cm (more than a foot) of space to crawl
under the ring. You could easily crawl under. If the ring is gathered as below then you'd
have the full 63.4cm.
3. The problem lies in that the wealthy Arab did not divide up his entire estate. 1/2 + 1/3 + 1/9
= 9/18 + 6/18 + 7/18 + = 17/18. The Arab has not accounted for 1/18th of his estate.
4. (a) If you saved 100% on fuel you wouldn't need any in order to travel - impossible.
(b) If a 100L was normally used then only 90L would be needed to get by the first device;
60% of the 90L would be used to get by the second device ie. 54L and 50% of the 54L or
27L would be used to to get by the third device. Thus the actual saving would be 73L of the
original 100L so there'd be a 73% saving.
5. (b) base = 10 units; altitude = 12 units, area = 1/2(10)(12)= 60u²
(d) base = 10 units; altitude = 12 units, area = 1/2(10)(12) = 60u².
There is a hole in the center whose area is 1(2) or 2u². This leaves an apparent area of 60 - 2
or 58u².
(e) The orignal figure is not really a triangle.
and
are not really line segments, that is,
A, D, and B are not collinear; likewise A, E, and C are not collinear.
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tan = 3/7
tan = 5/2
= tan-1(3/7)
= tan-15/2
o
= 23.19859051 = 68.19859051o
FDE = 90o
Thus there is a kink (dog's leg) as you travel from B to D to A. In exaggerated form it looks
like the figure below:
By Pathagorus AD = 29 and BD = 58
Therefore the area of ADB (the extra piece on the left) is 1/2
29 58 sin 178.602819o = .5 or 1/2u²
Similarly AEC hasan area of .5u²
Therefore the original figure has an area of 60u² - 2(.5u²) =
59u². You can verify this by adding together the areas of the
individual pieces.
Figure 2 has a kink the otherway - see at right, so the area we
got, 58u², is too small.
Note: + WQS + = tan-1(2/5) + 90o + tan-1(7/3) =
178.602819o
Therefore PQR has area 1/2( 29)( 58)sin178.602819o
Therefore area PQR = .5u². Likewise PST has area .5u²
Therefore the correct area of the second figure is 58u² plus the
2 mini 's each of area .5u² for a total of 59u²
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Lesson 2
Inductive Reasoning - Solutions
1. a) 1 = 1² ; 1 + 3 = 4 = 2²; 1 + 3 + 5 = 9 = 3²; 1 + 3 + 5 + 7 = 16 = 4²;
1 + 3 + 5 + 7 + 9 = 25 = 5²
b) 1 + 3 + 5 + 7 + 9 + 11 = 6²
1 + 3 + 5 + 7 + 9 + 11 + 13 = 7²
c) 10² or 100
d) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100 = 10²
e) yes
f) inductive reasoning is being used because we are forming a conclusion based on
observations.
2. (a) 2 colors
(b) 3 colors
(c) 3 colors
(d) 4 colors
(e) 4 colors
(f) that's impossible
(g) only four colors (maximum) are needed to color a map and distinguish the borders.
(h) No. Our conclusion was based only on our attempts in (f).
3. (e) answers will vary
(f) h= + t + u (see explanation below as to why.)
Area XYZ = Area PYZ + Area XPZ + XPY
1/2(YZ)h = 1/2(YZ) + 1/2(XZ)t + 1/2(XY)u (eq'n 1)
but YZ = XZ = XY, so dividing (eq'n 1) by 1/2YZ
we have h = + t + u
5.
(a) 23.64 = 1472
32.46 = 1472
69.64 = 4416
96.46 = 4416
26.93 = 2418
62.39 = 2418
84.36 = 3024
48.63 = 3024
41.28 = 1148
14.82 = 1148
(b) 53.71 = 3763 Therefore it is not true for all numbers
35.17 = 595 of 2 digits.
7. (c) For each of the pairs of numbers listed the product of the ten's digits is the same as the
product of the units digits. In 84.36 for example, (8)(3) = (4)(6). This can also be proven.
If the two numbers are of the form:
ab*cd
ba*dc
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8.
then (10a + b)(10c + d) = (10b + a)(10d + c)
100ac + 10ad + 10bc + bd = 100bd + 10bc + 10ad + ac
Therefore: 100ac + bd = 100bd + ac
99ac - 99bd = 0
ac - bd = 0
ac = bd
9. A fun exercise is to find all two digit numbers that have that property.
10.
(a) 8 9 10 11 12
113 131 151 183 197
20 21 .....
41
461 503
13 14 15 16
223 251 281 313
-
1763 =
17 18 19
347 383 421
41.43
11. Since f(x) = x² + x + 41, if x = 41, then
41²+ 41 + 41 = 41(41 + 1 + 1) = 41(43).
12. The program listed on the next page, can be used on the TI-82 to find the prime factors of a
number. It is pretty well essential if you want to do this problem in < 15 minutes.
13. PROGRAMD: PRM FCTRS
: Input "NMSBR?", N
:2-P
: Lb1 1
: If int (N/P) = N/P
: Goto 3
: IS > sss(P, N) : Goto 1 [IS - this is found underPRGM CTL. It means increment and
skip if greater than.]
: Goto 4
: Lbl 3
: N/P - N
: Pause P [pauses and prints the value of P.
: Goto 1
: Lbl 4
: If N 1 : Disp N
14. (b) Inductive reasoning might lead us to believe that x²+ x + 41 always results in a prime
number.
15. (c) Inductive reasoning suggests things that we may then be able to prove. It informs us of
the observed patterns but does not prove that the patterns always hold.
16. 220-1or 219 or 524288
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Lesson 3
Deductive Reasoning Solutions
1. To win you must take toothpick #1. You can do this if you pick up toothpick #5 (and stop
there). Likewise you need to pick up 13, and 17. So to guarantee a win, let your opponent
start. That way you'll be able to get #17 on your first turn. The sequence of "winning
toothpicks" is 17,13,9,5,1 whereas when you take 1 or 2 toothpicks the winning sequence
was 19,16,13,10,7,4,1.
2.
a) Choose any non zero number
b) Double the number
c) Subtract 1
d) Triple your answer
e) Add 3
f) Divide by the orginal number
x
2x
2x - 1
6x - 3
6x
6
10
20
19
57
60
6
3. If x matches are removed there will be 20-x matches remaining. Since 1 less than/equal to x
and x is less than/equal to 9, the ten's digit of 20-x must be a 1 and the units digit of 20-x
will be 10-x. Therefore, the sum of the digits of the number of matches left in the book will
be 1+10-x or 11-x. If 11-x matches are now removed there will only be 20-x-(11-x) or 9
matches left. Tearing out 2 more matches leaves 7.
Another proof (proof by exhaustion) may be easier for some students.
20
-1
19
- 10
9
-2
7
20
-2
18
-9
9
-2
7
20
-3
19
-8
9
-2
7
20
-4
19
-7
9
-2
7
20
-5
19
-6
9
-2
7
20
-6
19
-5
9
-2
7
20
-7
19
-4
9
-2
7
20
-8
19
-3
9
-2
7
20
-9
19
-2
9
-2
7
4. x²+ 8x + 7 = (x + 7)(x + 1). Provided x does not equal 0, in which case x²+8x+7 would give
7, x²+8x+7 will be factorable.
5.
a. By pulling out only two socks you can have 1 red and 1 white. By pulling out a third
sock you will definitely have a pair.
b. To get two pairs we know that 3 socks will certainly provide 1 pair leaving 1 extra
sock. The 4th sock may not match the extra sock but the 5th sock will match either
the 4th sock or the extra sock. Therefore 5 socks are required.
c. To get x pairs of socks, the minimum number of socks I would need would be 2x
socks. Now 2x is an even number of socks. If I had an even number of reds and an
even number of whites, then 2x socks could provide x pairs. However this 2x even
number of socks that are pulled out could consist of an odd number of reds and an
odd number of whites since two odd numbers also add to make an even number. If I
have an odd number of reds and an odd number of whites I won't be able to form x
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pairs. I will be able to form x-1 pairs having one extra red and 1 extra white. To
guarantee my final pair of socks, I will have to pull out 1 more sock for a total of
2x+1 socks. Therefore 2x+1 socks are required to guarantee x pairs.
6. If the person putting down the coins covers either 2 dark squares or two whites squares, the
person with the paper clips can't win. With seven paper clips, 14 squares will have to be
covered, 7 of which will be dark and seven of which will be white (each clip covers 1 dark
and 1 white square). Thus if the person with the coin has covered say 2 dark squares, the
board remains with 6 dark squares and 8 white squares making it impossible to cover 7 of
each color.
7. The three were grandfather, father and son.
8.
b + c = 106
b - c = 100
2b = 206
b = 103
Therefore c = 3
many think the botte costs $1 and
the cork costs $.06
9. It takes a cat 3 minutes to catch a mouse. Therefore 100 cats catch 100 mice in 3 minutes
also
10. If it were x's turn, x would win. Therefore it must be o's turn. Since there are two each
already marked, o must have started.
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Lesson 4
Conditional Statements and Proofs By Counter Example
1. hypothesis: "there's a will"
conclusion: "there's a way"
converse: If there's a way then there's a will.
Whether the converse is true is argueable. Judging by the actions of some people their way
through life shows little will. Many have no desire or goals. So is their will to have no will?
2. hypothesis: -5x+1 is less than 16
conclusion: x is greater than -3
converse: If x is greater than -3 then -5x is less than 16. Converse is true.
3. hypothesis:
are opposite rays.
conclusion: BAC is a straight angle.
converse: If BAC is a straight angle, then
are opposite rays. Converse is true.
4. hypothesis: you are registered in a math class
conclusion: you will need a calculator
converse: If you need a calculator, then you are registered in a math class. Converse is false.
5.
a) no conclusion possible
b) no conclusion possible
c) it does not have antilock brakes
7.
a) you are not my student
b) no conclusion possible
c) no conclusion possible
9.
a) I'll pick you up
b) It did not rain
c) no conclusion possible
d) no conclusion possible
11.
a) You don't own a Saturn
b) no conclusion possible
c) no conclusion possible
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12. 2 is prime but it's even
13. (-5)²= 5² but -5 5
14. 3/2
(3 + 1)/(2 + 1) scince 11/2 > 11/3
15. If j = -1 and k = 2 then -1/2
[2(-1)]/2 scince -1/2 > -1
16. There are no primes between 90 and 95, for example, since 91 = 13.7, 92 = 46.2, 93 = 31.3,
94 = 47.2.
17. First he took the duck across the river.
Then he took the fox across and returned with the duck.
Then he took the corn across and returned back to get the duck.
18. The man went to town on Tuesday. He couldn't have gone on Thursday or Friday because
the meat market was not open. He couldn't have gone on Saturday because the eye doctor
was closed. He must have been at the bank to cash his check since he came home with more
money than he left with. Thus he must have gone to town on Tuesday, the only remaining
day that the bank was open and the meat market and the eye doctor. (Assume that he
stopped on the same day he went to the bank.)
19. Suppose the boys name were Al, Ben, Carl. Suppose Al was the one to raise his hand. He
would have reasoned as follows. "If I don't have dirt on my head then Ben or Carl raised
their hands the first time because each of them saw the dirt on each other's head and didn't
see any on mine. If that were true then surely Ben or Carl would realize why the other had
raised his hand the first time and each would know that he had dirt on his head and would be
raising his hand the second time. Because neither Ben or Carl is raising their hand the
second time it must not be that I don't have dirt on mine. Therefore I do have dirt on mine."
20. His friend was a lady, not a man, and her name was Anne.
21. Since there ages multiplied to 36, they must have been 1,6,and 6 or 2,2,and 9. In each case
the sum of their ages is the same.(Any other possible combinations of ages, like 1,4,9 or
2,3,6 give unique sums.) Since the oldest girl did not like chocolate pudding their ages must
have been 2,2, and 9 because if they had been 1,6, and 6 there could not have been an oldest
girl.
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Lesson 5
Proofs About The Properties Of integers
1. Let the two odd integers be b and c
Therefore b = 2k+1 and c = 2j+1 where k and j are integers
Therefore b + c = 2k+1 = 2j+1
= 2k + 2j + 2
= 2(k + j) + 1
Therefore b + c is odd since 2(k+j)+1 is even
2. Let the even integer be b
Let the odd integer be c
Therefore b = 2k and c = 2j + 1
Therefore b + c = 2k + 2j + 1
= 2(k + j) + 1
Therefore b + c is odd since 2(k + j)+ 1 is odd
3. Let the two odd integers be b and c
Therefore b = 2k + 1
Therefore c = 2j + 1
bc = (2k + 1)(2j + 1)
= 4kj + 2k + 2j + 1
Therefore bc is odd since 2(2kj + k + j) + 1 is the form of an odd integer
4. Let two consecutive integers be a and a + 1.
Now if a is even, then a = 2k and a + 1 = 2k + 1
Therefore a(a + 1) = 2k(2k + 1) = 4k²+ 2k = 2(2k²+ k) which is even.
Now if a is odd, then a = 2j + 1 and a + 1 = 2k + 1
Therefore a(a + 1) = 2k(2k + 1) = 4k²+2k = 2(2k²+ k) which is even.
Now if a is odd, then a = 2j + 1 and a + 1 = 2j + 2
Therefore a(a + 1) = (2j + 1)(2j + 2) = 4j²+ 4j + 2j + 2
= 4j²+ 6j + 2
= 2(2j²+ 3j + 1) which is even.
Therefore the product of 2 consecutive integers will always be even.
5. Let the three consecutive integers be x, x + 1, x + 2.
Therefore their sum = x + x + 1 + x + 2
= 3x + 3
= 3(x + 1) which is a multiple of 3.
6. Suppose t is the common factor of x and y.
Then x = tk and y = tj
Therefore x + y = tk + tj
= t(k + j)
Therefore t is a common factor of their sum.
7. Let the even integer be c.
Therefore c = 2k
Therefore c² = (2k)(2k) = 4k² = 2(2k²) which is even in form.
8. Let the two consecutive integers be x and x + 1
Therefore (x + 1)2 - x² = x² + 2x + 1 - x² = 2x + 1 which is in odd form.
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9. Let the two odd integers be x and x + 2
Therefore x² + (x+2)² = x² + x² + 4x + 4 = 2x² + 4x + 4 = 2(x² + 2x + 2) which is fo even
form.
10. Let the ten's digit be t. Since the number ends in 5 it can be written as 10t + 5.
therefore (10t + 5 )2 = 100t2 + 100t + 25
= 100t(t + 1) + 25
where t = 10's digit and t + 1 is next larger digit
11. Let the two digit number be 10t + u
The number with digits reversed is 10u + t
10t + u - (10u + t)
= 10t + u - 10u - t
= 9t - 9u
9(t - u) which is of the form to e a multiple of 9.
12. Let x be the integer. If x is even, then x = 2k.
Therefore x² = (2k)² = 4k² which is a multiple of 4. If x is odd then x = 2k + 1. Therefore x²
= (2k + 1)² = 4k² + 4k + 1 = 4(k² + k) + 1 which is of the form to be 1 more than a multiple
of 4.
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Lesson 6
Deductive Geometric Proofs and Indirect Proofs
1. Assume that
. Scince
bisects
then
. Also QWB since they are
vertical angles. Thus AWP
QWB by SAS. Therefore
(corresponding parts
of
's). This contradicts the given fact that
. Therefore our assumption is false.
therefore
2. Assume that
. Since
and
then WXY
ZYX since both are
right angles. Also
(common side) so WXY
ZYX (S.A.S.). Therefore
(corresponding parts of
's). This contradicts the given fact that
. Therefore
our assumption is false:
3. Assume that
. Now
(given) so S
W (In STW if
then
the angles opposite these sides are also ). Therefore STX
WTY (S.A.S.).
Therefore
(corres. parts of
's). This contradicts the given fact that
.
Our assumption is false:
4. Assume that
. Now
(given) so S
W(
's lie opposite sides
in a .) Therefore SXT
WYT (S.A.S.) so SXT
WYT (corres. parts of
).
Therefore TXY
TYX (they supplement
's SXT and WYT since SXY and
WYX are straight angle's. This contradicts the given fact that TXY
TYX. Therefore
our assumption is false:
5. Assume that there is a largest integer k.
Since the integers are closed under addition (that is the sum of any two integers is also an
integer) then k + 1 is an integer.
Since k is the largest integer, then k + 1 < k
Therefore 1 < 0 (subtract k form each side)
Clearly 1
0. Therefore our assumption was false: thereis no largest integer.
6. Assume that 3 is a rational number. Then 3 = a/b where a and b are integers with no
common factor.
Therefore a = 3b
Therefore a²= 3b²
Therefore a² is a multiple of 3. If a² is a multiple of 3, then so is "a" a multiple of 3. (only
multiples of 3 have squares that are multiples of 3. Consider 1², 2², 3², 4², 5², 6², 7², etc. only
3², 6², 9², etc. are multiples of 3). If "a" is a multiple of 3 then "a" = 3w where w is an
integer. Therefore since a²=3b², then (3w)²=3b² or 9w²=3b² so 3w²=b². Therefore b² is a
multiple of 3 and thus b is a multiple of 3. Therefore both a and b are multiples of 3. This
contradicts the fact that if 3 is rational "a" and b have no common factor. Therefore our
assumption is false. Therefore 3 is rational.
7. Assume that the product of two even integers is odd. Then if "a" and "b" are the even
integers, ab = 2k + 1 where k is an integer. Since "a" and "b" are even integers then a = 2w
and b = 2j where w and j are integers. Therefore:
ab = (2w)(2j) = 2k + 1
therefore 4wj = 2k + 1
4wj - 1 = 2k
(4wj - 1)/2 = k
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2wj - 1/2 =k
Since 2, w, and j are integers, so is 2wj. But 2wj-1/2 could not be an integer. This
contradicts the fact that k is an integer. Therefore our assumption is false. Therefore the
product of two even integers is even.
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Lesson 7
Coordinate Geometry Proofs
1.
2.
3.
4.
5.
6.
7.
8.
C(a,a); D(0,a)
B(a,0); D(0,b)
C(a + b, c)
C(b,c) or any two other variaaables that are not the same as "a" or each other.
C(a - b, c)
C(b,c); D(d,e)
C(a + b, c)
Let the coordinates of the vertices of the rectangle be as shown:
AC = [(a - 0)² + (b - 0)²] = (a² + b²)
BD = [(0 - a)² + (b - 0)²] = (a² + b²)
Therefore AC = BD
9. Let the coordinates of the be as shown:
M's coordinats are: [(b + 0)/2 , (c + 0)/2] = [b/2 , c/2]
N's coordinats are: [(b + 0)/2 , (c + 0)/2] = [(b + a)/2 , c/2]
Therefore MN has a slope of {[c/2 - c/2]/[(b + a)/2 - b/2]} =
0/(a/2) = 0
Therefore MN // BC
The length of BC = a. The length of MN = {[(b + a)/2]² +
[c/2 - c/2]²}
= [(a/2)² + 0 = a² = a/2 Therefore MN's lenght is half of
BC's lenght
10. Let the coordinates of the rhombus be as shown:
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Slope for
= (c - 0)/(a + b - 0) = c/(a + b)
Slope for
= (c - 0)/(a - b - 0) = c/(b - a)
We must show that [c/(a + b)][c/(b - a)] = -1
11.
Since ABCD is a rhombus we know that AB + BC
Therefore sq. root of [(a - o)² + (0 - 0)²] = sq. root [(a + b - a)² + (C - 0)²
a² = b² + c²
Then [c/(a + b)(c/(b - a)]=c²/(b² - a²) = 9a² - b²)/(b² - a²) = -1²
Therefore they (
and
) are perpendicular.
12. Let the coordinates of the parallelogram be as shown
We know that AC + DB
therefore [(a + b - 0)² + (c - 0)²] = [(a + b)² + (c - 0)²]
therefore (a + b)² + c² = (b - a)² + c²
therefore a² + 2ab + b² + c² = b² - 2ab + a² + c²
therefore 2ab = -2ab
therefore 4ab = 0
therefore a = 0 or b = 0
13.
a can't be 0 because then B would coincide with A and the figure could not be a
parallelogram. (there is no figure)
therefore b must be 0. this means D lies on the axis. Therefore DAB = 90o
Therefore ABCD is a rectangle since a parallelogram having 1 right angle is a rectangle.
14. Let the coordinates of the trapezoid (isosceles) be as shown.
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AC + [(a - b - 0)² + (c - 0)²]
= [a² - 2ab + b² + c²
BD = [(b - a)² + (c - 0 )²]
= [b² - 2ab + a² + c²]
Therefore AC = BD
15. Let the triangle have vertices as shown.
We need to sow that
. We will try to show that the
product of the slopes of
and
is -1.
Slope of
= (c - 0)/(b - 0) = c/b
Slope of
= (c - 0)/(b - a) = c/(b - a)
Therefore (Slope of
)(Slope of
= [c/b][c/(b - a)] =
[c²/b(b - a)] {eq'n 1}
16.
M, the midpoint of
, has coordinates of (a/2, 0)
Since AM + MC, then [(a/2 - 0)² + (0 - 0)²] =
[(b - a/2)² + (c - o)²]
a²/4 = (b - a/2)² + c²
a²/4 = b² - ab + a²/4 + c²
0 = b² - ab + c²
therefore c = c² {eq'n 2}
Going back to eq'n 1 we have c²/(b(b - a))= (ab - b²)/(b(b - a)) = (b(a - b ))/(b(b - a)) = -1
Therefore ABC is a right angle.
17. Let the trapazoid have coordinates as shown.
x has coordinates [(d + 0)/2, (c + )/2]=(d/2, c/2)
y has coordinates [(b + a)/2, (c + )/2]=((b + a)/2, c/2)
therefore slope of
= [(c/2 - c/2)/((b + a)/2 - d/2) = 0/((b +
a)/2 - d/2) = 0
Therefore
//
since
's slope is also 0.
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18.
AB = [(a - 0)² + (0 - 0)²] = a
XY = [((b + a)/2 - d/2)² + (c/2 - c/2)²] = (b + a)/2 - d/2 = (b + a - d )/2
DC = [(b - d)² + (c + c)²] = b - d
1/2(AB + DC) = 1/2(a + b - d) = (a + b - d)/2
* subtract coordinates so differences remain positive. Otherwise use absolute values when
you take the .
19. Let the quad. have coordinates as shown:
Slope of
= [(f + d)/2 - d/2]/[(c + e)/2 - (c + b)/2]
= [f/2]/[e/2 - b/2] = f/(e - b)
Slope of
= [f/2 - 0]/[e/2 - b/2]
= [f/2]/[e/2 - b/2] = f/(e - b)
20.
Therefore
//
. Similarly one can show that the slope of
are therefore //. Therefore Quad XYZW is a llgm
https://www.k12.gov.sk.ca/docs/math30/app-i.html
and
are both d/c and
Lesson 8
Proofs By Mathematical Induction
1. Is the statement true when n is say 3?
Basis Step
Does 1 + 2 + 3 =? [3(3 + 1)]/2
6 =? [3(4)]/2
Yes
2.
Does
Assume the statement is true for n = k
ie assume that 1 + 2 + 3 + 4 +...+ k = [k(k + 1)]/2. Now prove that the statement is true when
n=k+1
That is prove that 1 + 2 + 3 + 4 + ....+ k + k + 1 = (k + 1)(k + 1 + 1)]/2
L.S. (assumption) = k(k + 1)/2 + k + 1
= [k(k + 1) + 2(k + 1)]/2
= (k + 1)(k + 2)/2
R.S. = (k + 1)(k + 2)/2
3.
since L.S. = R.S. we have proven the statement is true when n = k + 1. Therefore the
statement is true for all natural numbers.
4. Is the statement true when n = 3?
Basis Step
Does 2 + 6 + 10 =? 2(3)²
18 =? 2(9)
18 = 18 Yes
5. Assume that the statement is true for n = k
ie assume that 2 + 6 + 10 + 14 +...+ 4k - 2 + 4(k + 1) - 2 = 2(k + 1)² Now prove that the
statement is true for n = k + 1
That is, show that 2 + 6 + 10 + 14 +...+ 4k - 2 + 4(k + 1) - 2 = 2(k + 1)²
L.S. assumption = 2k² + 4k + 4 - 2
= 2k² + 4k + 2
= 2(k² + 2k +1)
= 2(k + 1)²
6.
L.S. = R.S. therefore true for all natural numbers
7. Is the statement true for say n = 3?
Does 1 + 2 + 4 = 23 - 1 ?
7 = 7 Yes.
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8. Assume that the statement is true for n = k
ie assume that 1 + 2 + 4 + 8 +...+ 2k - 1 = 2k - 1
Now prove that the statement is true for n = k + 1
ie Prove that
1 + 2 + 4 + 8 +...+ 2k - 1 + 2k + 1 -1 = 2k + 1 - 1
= 2k - 1 + 2k
= 2k + 2k - 1
= 2(2k) - 1
= 2k + 1 - 1
9.
L.S. = R.S. therefore true for all natural numbers
10. Is the statement true when n = 3?
Does 1² + 2² + 3² = [3(3 + 1)(2(3) + 1)]/6 ?
1 + 4 + 9 =? 3(4)(7)/6
14 = 84/6
14 = 14
11. Assume that the statement is true for n = k
ie assume that 1² + 2² + 3² + 4² +...+ k² = k(k + 1)(2k + 1)/6
Prove that the statement is true for n = k + 1 ie Prove that 1² + 2² + 3² + 4² +...+ k² + (k + 1)²
= (k + 1)(k + 1 + 1)(2(k + 1) + 1)/6
L.S. = k(k + 1)(2k + 1)/6 + (k + 1)² R.S. = (k + 1)(2k + 3)(k + 2)/6
= [k(k + 1)(2k + 1) + 6(k + 1)²]/6
= (k + 1)[k(2k + 1) + 6(k + 1)]/6
= (k + 1)[2k² + k + 6k + 6)/6
= (k + 1)(2k² + 7k + 6)/6
= (k + 1)(2k + 3)(k + 2)/6
12.
L.S. = R.S. therefore true for all natural numbers
13. Is the statement true when n = 3?
Does 1² + 3² + 5² = [3(2(3) + 1)(2(3) - 1)]/3 ?
1 + 9 + 25 =? 3(7)(5)/3
35 = 35
14. Assume that the statement is true for n = k
ie assume that 1² + 3² + 5² +...+ (2k - 1)² = k(2k + 1)(2k - 1)/3
Prove that the statement is true for n = k + 1 ie Prove that 1² + 3² + 5² +...+ (2k - 1)² + (2(k +
1))² = (k + 1)(2(k + 1) + 1)(2(k + 1) - 1)/3
L.S. = k(2k + 1)(2k - 1)/3 + (2k + 1)² R.S. = (k + 1)(2k + 3)(2k + 1)/3
= [k(2k + 1)(2k - 1) + 3(2k + 1)²]/3
= (2k + 1)[k(2k - 1) + 3(2k + 1)]/3
= (2k + 1)[2k² - k + 6k + 3)/3
https://www.k12.gov.sk.ca/docs/math30/app-i.html
= (2k + 1)(2k² + 5k + 3)/3
= (2k + 1)(2k + 3)(k + 1)/3
15.
L.S. = R.S. therefore true for all natural numbers
16. Is the statement true when n = 4?
Is 4² + 4/2 ?
Is 20/2 ? Yes
Assume that the statement is true when n = k.
That is assume that k² + k is divisible by 2.
Prove that the statement is true for n = k + 1. That is, prove that (k + 1)² + (k + 1) is divisible
by 2
(k + 1)² + (k + 1) = k² + 2k + 1 + k + 1 = k² + k + 2k + 2
= k² + k + 2(k + 1)
[k² + k] is div. by 2 by assumption
[2(k + 1)] is div. by 2 by form
Therefore sum is divisible by 2.
Therefore true for all natural numbers.
17. Is the statement true when n = 4?
ie is 43 + 2(4) divisible by 3?
64 + 8 = 72 which is divisible by 3.
Assume that the statement is true when n = k.
ie assume that k3 + 2k is divisible by 3.
Prove that the statement is true for n = k + 1.
ie prove that: (k + 1)3 + 2(k + 1) is divisible by 3
= k3 + 3k² + 3k + 1 + 2k + 2
= k3 + 2k + 3k² + 3k + 1 + 2
= k3 + 2k + 3k² + 3k + 3
= k3 + 2k + 3(k² + k + 1)
[k3 + 2k] is a multiple of 3 by assumption
[3(k² + k + 1)] is div. by 2 by form
18.
Therefore sum is a multiple of 3.
Therefore true for all natural numbers.
19. Is the statement true for n = 5?
Is 5(5² + 5) divisible by 6?
Is 5(30) divisible 6?
Is 150 divisible by 6? Yes
Assume that the statement is true for n = k
ieassume that k(k² + 5) is divisible by 6
https://www.k12.gov.sk.ca/docs/math30/app-i.html
Now show that the statement is true for n = k + 1
ieshow that: (k +1)((k + 1)² + 5) is divisible by 6
= (k +1)(k² + 2k + 1 + 5)
= (k +1)((k² + 5 + 2k + 1)
= k(k² + 5) + k(2k + 1) + 1(k² + 5) + 1(2k + 1)
= k(k² + 5) + 2k² + k k² + 5 + 2k + 1
= k(k² + 5) + 3k² + 3k + 6
= k(k² + 5) + 3(k(k + 1) + 2)
[k(k² + 5)] is a multiple of 6 by assumption
[3(k(k + 1) + 2)] is multiple of 3 by form, but since k and k + 1 are consecutive integers,
their product will be even. an even number + 2 is still even. Therefore k(k + 1) + 2 is even.
Therefore3(k(k + 1) + 2) is also a multiple of 6.
20.
Therefore their sum will be a multiple of 6
21. Is the statement true when n = 2?
Is 2 < 2 + 1 ? Yes
Assume that the statment is true when n = k.
That is, assume that k < k + 1
Is the statement true when n = k + 1?
ie Prove that k + 1 < k + 1 + 1 ie (k + 2)
Since k < k + 1 (by assumption)
and 1 = 1
then k + 1 < k + 1 + 1 (+ n of 1 to both sides).
Therefore k + 1 < k + 2.
Therefore true when n = k + 1.
Therefore true for all natural numbers.
22. Is the statement true when n = 3?
is 2 < 23 ? Yes.
Assume that the statement is true for n = k.
ie assume that 2 < 2k
Now prove that the statement is true for n = k + 1.
ie Prove that: 2 < 2k + 1
If 2 < 2k (by assumption)
then 2·2 < 2·2k
therefore 4 < 2k + 1 (mult. by 2)
but 2 < 4
therefore 2 < 2k + 1
Therefore the statement is true for n = k + 1.
Therefore true for all natural numbers.
23. Is the statement true when n = 6?
ie is 26 > 6²
Assume that the statement is true n = k.
ie assume that 2k > k²
Now prove that the statement is true for n = k + 1
ie prove that 2k + 1 > (k + 1)² ie prove 2k + 1 > k² + 2k + 1 (eq'n 3)
If 2k > k² (eq'n 1)(by assumptiion)
then 2k + 2k > 2k + k² (addition of 2k to both sides of eq'n 1)
https://www.k12.gov.sk.ca/docs/math30/app-i.html
therefore 2(2k) > 2k + k²
therefore 2k + 1 > 2k + k² (eq'n 2)
Now 2k + k² will be > k² + 2k + 1 if 2k is > 2k + 1. this is true if k = 2 and so will certainly
be true for any n value since n > 5.
Therefore since 2k + 1 > 2k + k² (eq'n 2) AND 2k + k² > k² + 2k + 1,
then 2k + 1 > k² + 2k + 1 which is what we needto prove in eq'n 3.
Therefore true for all natural numbers.
24. Is the statement true when n = 3?
Is 2(3) < 23 ?
6 < 8 Yes
Assume that the statement is true when n = k.
ie assume that 2k < 2k.
Now prove that the statement is true when n = k + 1
That is, prove that 2(k + 1) < 2k + 1 (eq'n 3)
Since 2k << 2k (a assumption)
then 2(2k) < 2(2k) (mult. by 2)
4k < 2k + 1 eq'n 1
How does 4k compare to 2(k + 1)
4k - 2(k + 1) = 2k - 1 which is positive since k > 1
Therefore is 4k - 2(k + 1) is positive then 2(k + 1) < 4k eq'n 2
Combining eq'n 1 and eq'n 2 we have
2(k + 1) < 2k + 1 which proves eq'n 3 is true for all natural numbers
25. Is the statement true when n = 2?
Is x² - y² is divisible by x - y?
Since (x - y)(x + y) = x² - y², then x² - y² is divisible by x - y.
Assume that the statement is true when n = k.
ie assume that xk- yk is divisible by x - y.
Now prove that the statement is true when n = k + 1.
ie prove that xk + 1 - yk + 1 is divisible by x - y.
Now xk + 1 - yk + 1
= xk + 1 - xyk + xyk - yk + 1 but since(-xyk + xyk = 0)
= x(xk - yk) + yk(x - y)
[(xk - yk)] is div. by (x - y) by assumption
[yk(x - y)] is div. by (x-y) since (x - y) is a factor
Since each term is divisible by x-y so is their sum. Therefore the statement is true when n =
k + 1.
Therefore true for all natural numbers.
26. Is the statement true when n = 3?
Does a + (a + d) + ( a+ 2d) = 3/2[2a + (3 - 1)d] ?
Does 3a + 3d = 3/2[2a + 2d] ?
3a + 3d = 3(a + d) Yes
27. Assume that the statement is true when n = k. ie assume that a + (a + d) + ( a+ 2d) +...+ ( a +
(k - 1)d) = k/2[2a + (k - 1)d]
Prove that the statement is true when n = k + 1.
ie prove that a + (a + d) + ( a+ 2d) +...+ ( a + (k - 1)d) + (a + (k + 1 - 1)d) = k/2[2a + (k 1)d]
https://www.k12.gov.sk.ca/docs/math30/app-i.html
L.S. = k/2[2a + (k - 1)d] + (a + (k + 1 - 1)d) R.S. = (k + 1)/2[2a + (k + 1 - 1)d]
= k/2[2a + (k - 1)d] + 2(a + kd)/2
= (k + 1)/2[2a + kd]
= 1/2{k[2a + (k - 1)d] + 2(a + kd)}
= 1/2[(k + 1)(2a +kd)]
= 1/2{k[2a + kd - d] + 2a + 2kd)}
= 1/2[2ka + k²d + 2a + kd]
= 1/2[2ka + k²d - kd + 2a + 2kd]
= 1/2[2ka + k²d + 2a + kd]
28.
L.S. = R.S. therefore true for all natural numbers
29. Is the statement true when n = 3?
ie does a have interior whos sum is (3 - 2)180 or 180? Yes.
Assume that the statement is true when n = k.
ie assume that a convex polygon of k sides has an interior angle sum of (k - 2)180.
It is true for n = k + 1
ie Prove that the interior angles of a convex polygon of (k + 1) sides have a sum of (k + 1 2)180o.(eq'n 1)
A polygon of k sides permits one to draw k - 3 interior diagonals from a particular vertex,
say A because connecting A to itself or to its adjacent vertices (one per side) does not create
interior diagonals. By drawing k - 3 diagonals the polygon is divided into k - 2 's each of
whose sums is 180o hence giving the (k - 2) 180o total. When the k + 1 st vertex is created to
give the k + 1 st side then A can be connected to one or more vertex resulting in the creation
of one or more whose interior have a sum of 180o. Therefore a polygon of k + 1 sides
will have an interior sum of (k - 2)180o + 180o = 180(k - 2 + 1)(eq'n 2)
[(k - 2)180o] is the sum for a k sides
[180o] is the increase for k + 1 st side
Since eq'n 1 = eq'n 2 the statement is true when n = k + 1. Therefore true for all natural
numbers (> 3)
30. Is the statement true when n = 3. ie do 3 lines drawn through a given point divide the plane
into six regions?
Yes
31.
32. Assume the statement is true when n = k. ie assume that k lines that pass through P will
divide the plane into 2k regions. Prove that the statement is true when n = k + 1. ie prove
that the plane will be divided into 2(k + 1) regions by k + 1 lines. Since k lines divide the
plane into 2k regions. When 1 or more line is drawn through P it will be cut through two
existing regions creating 2 more regions (each of those two existing regions is cut into two
making 4 regions where before there were 2). Therefore there are now 2k + 2 regions or 2(k
+ 1) regions which is what we had to show. Therefore the statement is true when n = k + 1.
Therefore it is true for all natural numbers.
33. Is the statement true when n = 5. Can a postage of $.05 be made up using two cent and three
cent stamps? Yes $.05 = 1($.02) + 1($.03) Assume that the statement is true when n = k. ie
assume that a postage of k can be made up using two cent and three cent stamps.
https://www.k12.gov.sk.ca/docs/math30/app-i.html
If k is odd, then k + 1 will be even. Therefore k + 1 can be made using 1/2(k + 1) two cent
stamps. If k is even then k + 1 will be odd and k - 2 will be even. The k - 2 can be made
up of 1/2(k - 2) two cent stamps by using 1 three cent stamp in addition to the 1/2(k - 2) two
cent stamps you will make up your postage of k + 1 cents. Therefore the statement is true
when n = k + 1. Therefore it is true for all natural numbers.
34. Is the statement true when n=3? Does cos3 = (-1)3? yes
cos3 = -1 = (-1)3
35.
Assume that the statement is true when n = k. ie assume that cos k = (-1)k Prove that the
statement is true when n = k + 1.
ie prove cos (k + 1) = (-1)k + 1
Now as (k + 1) = cos (k + ) = cos k cos - sin k sin
= (-1)k(-1) - (sin k )(0)
= (-1)k(-1) - 0
= (-1)k
36.
37. Therefore statement is true when n = k + 1. Therefore it is true for all natural numbers.
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