1. a C was a knight. The argument runs as follows. If you ask a knight what

C H A P T E R2

Proofs Exercises 2.3

• (a) C was a knight. The argument runs as follows. If you ask a knight what

he is he will say he is a knight. If you ask a knave what he is he is obliged to lie and so also say that he is a knight. Thus no one on this island can say they are a knave. This means that B is a knave. Hence C was correct in saying that B lies and so C was a knight.(b) A is a knave, B a knight and C a knave. The argument runs as follows. If all three were knaves then C would be telling the truth which contradicts the fact that he is a knave. Thus at least one of the three is a knight and at most two are knights. It follows that C is a knave. Suppose that there were exactly two knights. Then both A and B would be knights but they contradict each other. It follows that exactly one of them is a knight.Hence A is knave and B is a knight.

• Sam drinks water and Mary owns the aardvark. The following table shows all

the information you should have deduced.

1 2 3 4 5

HouseYellowBlue Red White Green Pet FoxHorse Snails Dog Aardvark Name Sam Tina Sarah Charles Mary DrinkWaterTea Milk Orange juiceCoffee CarBentleyChevyOldsmobile Lotus Porsche The starting point is the following table.12345 House Pet Name Drink Car 1 Solutions Manual for Algebra & Geometry An Introduction to University Mathematics, 2e by Mark Lawson (All Chapters, 100% Original Verified, A+ Grade , Ch 1 Missing) 1 / 4

2Solutions to Algebra & Geometry: Second Edition

Using clues (h), (i) and (n), we can make the following entries in the table.

1 2 345

House Blue Pet NameSam Drink Milk Car There are a number of different routes from here. I shall just give some exam- ples of how you can reason. Clue (a) tells us that Sarah lives in the red house.Now she cannot live in the rst house, because Sam lives there, and she cannot live in the second house because that is blue. We are therefore left with the following which summarizes all the possibilities so far.

1 2 3 4 5

HouseYellow? White? Green?BlueRed?Red?Red?Pet Name SamSarah?Sarah?Sarah?DrinkMilk Car Clue (b) tells us that Charles owns the dog. It follows that Sam cannot own the dog. We are therefore left with the following possibilities.

12 3 4 5

House Yellow? White? Green?BlueRed?Red?Red?PetFox? Horse? Snails? Aardvark?NameSamSarah?Sarah?Sarah?DrinkMilk Car Both Questions 1 and 2 demonstrate some of the logic needed in mathematics.

3.other cases are proved similarly. Letmbe even andnbe odd. Thenm=2rand n=2s+1 for some integersrands. Thusm+n=2r+2s+1=2(s+r) +1.Thusm+nis odd.

4.The sum of the interior angles of the gure is equal to the sum of the angles in the two triangles. This is 360

.

5.p 2 2 +3 2 +7 2 = p

62. 2 / 4

Proofs3 (b) that diagonal. This has twice the area by Pythagoras' theorem.(c) 1 2 xybut we are told that it equals 1 4 z 2 . Hence 1 2 xy= 1 4 z 2 . By Pythagoras' theoremz 2 =x 2 +y 2 . Substituting this value forz 2 we get that 1 2 xy= 1 4 (x 2 +y 2

• Rearrange this to get(xy)

2 =0.Hencexy=0 and sox=y. It follows that the triangle is isosceles.

6.ncan be writtenn=10q+rwhere 0r9. It follows thatn 2 =10(10q 2 +2qr)+r 2 . Thus the last digit ofn 2 is equal to the last digit ofr 2 . Direct calculation now shows that this can only be one of 0;1;4;5;6;9. The converse is not true because 14 is a natural number ending in 4 but is not a perfect square.(b)nis even then its last digit is even and (2) if the last digit ofnis even thennis even. We prove (1). We may writen=10q+r. Thusr=n10q. Butnis even and 10qis even so it follows thatris even. We prove (2). We may writen=10q+rwherer is even. Thusr=2s. It follows thatn=10q+2s. It is now clear thatnis even.(c)nis divisible by 9 then the sum of the digits innis divisible by 9 and (2) if the sum of the digits innare divisible by 9 thennis divisible by 9. The proof is based on the following observation. Each power of 10 leaves remainder 1 when divided by 9. Thus 10=9+1 and 100=99+1 etc. It follows thatncan be written as a multiple of 9 plus the sum of its digits. The proofs of (1) and (2) now follow easily.

7.p

• is irrational,

except that you need to prove the following result: if 3 dividesa

2 then 3 divides

• To prove this we need the following. We may writea=3q+rwherer=

0;1;2. It follows thata 2 =3(3q 2 +2r)+r. Thusa 2 is divisible by 3 if and only ifais divisible by 3.

8.AˆDB=AˆCBusing Proposition III.21. Deduce that trian- gleAXDis similar to triangleABC. Show that the anglesDˆCA=DˆBAusing Proposition III.21. The anglesDˆAC=BˆAXfrom what we already know and how the angles are formed. Thus the trianglesAXBandACDare similar. The two equations follow immediately from the fact that we have two sets of simi- lar triangles. The remainder of the proof is just routine algebra using these two equations where we eliminatee.

9.y=tx+sfor sometand

• Our lines passes through the point(1;0)and sot=s. Thus it has the

formy=t(x+1).(b)P, we have to solve the equation x 2 +t 2 (x+1) 2 =1 forx. This yieldsx= 1t 2 1+t 2since we are excludingx= 3 / 4

4Solutions to Algebra & Geometry: Second Edition

• Substitute this value into 1=x

2 +y 2 and we obtain the corresponding valuey= 2t 1+t 2.(c) tis rational then the corresponding point(x;y)is rational. If(r;s)is a rational point on the unit circle then there is a corresponding rational value oftthat yields that point sinces=t(r+1).(d)(1;0)and all points of the form

1t 2 1+t 2; 2t 1+t 2

wheretis any rational number.With a little more work, it is possible to use this result to derive a formula for all Pythagorean triples.

10.cos q 2 = r 1+cosq 2 ; where 0q p 2 , and the fact that cos45

= 1 p 2 . An isosceles triangle with side 1 and enclosed angleqhas base p 2 p 1cosq. We may now calculate the lengths of the sides of a square (4-gon), octogon (8-gon), 16-gon, 32-gon etc as follows.n 4 8 1632 side p 2 p 2 p 2 q 2 p 2+ p 2 r 2 q 2+ p 2+ p 2 It is clear this pattern continues and it is now easy to derive our approximations top. This result suggests, but does not prove, thatpis a special kind of number.See Section 7.10.

11.Collatz problemor the 3x+1problem. Nobody has yet found a proof. It is therefore conceivable that there is a number where the process described in the question does not terminate. I included this question to show that unsolved problems are not limited to what you might regard as advanced mathematics.

• / 4