Direct Proof and Proof by Contrapositive
PROOF ANALYSIS
Result 3.6 Proof
PROOF ANALYSIS
Result 3.7
P roof
be necessary to establish the truth o f som e other m athem atical statem ents along the way that can then be used to establish the truth o f Q(n) . We w ill see exam ples o f this later.
L e t’s consider another exam ple. F or variety, w e use an alternative opening sentence and different sym bols in the p ro o f o f the follow ing result.
I f n is an even integer, then —5m - 3 is an od d integer.
L et n be an even integer. T h en n = 2x, w here x is an integer. Therefore,
—5 n - 3 = - 5 ( 2 x ) - 3 = - l O x - 3 = - l O x - 4 + 1 = 2 ( - 5 x - 2) + 1.
Since 5x - 2 is an integer, - 5 / ; - 3 is an odd integer. a We now consider another exam ple, w hich m ay have a surprise ending.
I f n is an od d integer, then 4 n 3 + 2n — 1 is odd.
A ssum e that n is odd. T hen n — 2_v + 1 for som e integer y . Therefore, 4 n 3 + 2 n - l = 4 (2 y + l )3 + 2(2y + 1) - 1
= 4 (8 y 3 + 12y2 + 6 y + 1) + 4 y + 2 - 1
= 3 2 y 3 + 4 8 y 2 + 28y + 5
= 2 (1 6 v 3 + 2 4 y 2 + 14y + 2) + 1.
Since 16y3 + 2 4 y 2 + 14y + 2 is an integer, 4 n 3 + 2n - 1 is odd. ■
A lthough the direct p ro o f o f R esult 3.6 th at w e gave is correct, this is not the desired proof. Indeed, had w e observed that
4n 3 + 2n - 1 = 4 n 3 + 2n - 2 + 1 = 2 (2 n 3 + n - 1) + 1
and that 2n 3 + n — 1 6 Z, w e could have concluded im m ediately that 4 n 3 + 2n — 1 is odd for every integer n. H ence a trivial p ro o f o f R esult 3.6 could be given and, in fact, is preferred. The fact that 4 //3 + 2 n - 1 is odd does not depend on n being odd. Indeed, it w ould be far b etter to replace the statem ent o f R esult 3.6 by
I f n is an integer, then 4/?3 + 2n - 1 is odd. +
W e give an additional exam ple o f a som ew hat different type.
t t c ii i -ii a i , c U w( « + 3 ) . ( n + 2 ) ( n - 5 ) . L e t S = {1, 2, 3} an d let n € S. I f --- --- is even, t h e n --- is even.
** 2
L et n e S such that n(n + 3 )/2 is even. Since n ( n + 3 ) /2 = 2 w hen n = 1, n(n + 3 ) /2 = 5 w hen n = 2 and n (n + 3 ) /2 = 9 w h e n /i = 3, it follow s that n = 1. W hen n = 1 ,(« + 2) (n - 5 ) /2 = - 6 , w hich is even. Therefore, the im plication is true. m
PROOF ANALYSIS In the p ro o f o f R esult 3.7, w e w ere only concerned w ith those elem ents n e S for w hich n(n + 3 )/2 is even. F urtherm ore, it is not initially clear for w hich elem ents n o f S the integer n (n + 3 )/2 is even. Since S consists only o f three elem ents, this can be determ ined
84 Chapter 3 Direct Proof and Proof by Contrapositive
rather quickly, w hich is w hat w e did. We saw that only n — 1 has the desired property
and this is the only elem ent w e n eeded to consider. ♦
If our goal is to establish the truth o f P ( x ) =>■ Q{ x ) for all x in a dom ain S by m eans o f a direct proof, then the p ro o f begins by assum ing that P ( x ) is true for an arbitrary elem ent x e S. It is often com m on in this situation, how ever, to om it the initial assum ption that P ( x ) is true for an arbitrary elem ent x e S. It is then understood that w e are giving a d irect proof. We illustrate this w ith a short exam ple.
Result 3.8 I f n is an even integer, then 3n 5 is an even integer.
P roof Since n is an even integer, n = 2 x for som e integer x . T herefore, 3 n 5 = 3 (2 x )5 = 3 (3 2 x 5) = 9 6 x 5 = 2 (4 8 x 5).
S ince 4 8 x 5 e Z, the integer 3n 5 is even. ■
F or the present, w hen giving a direct p ro o f o f P ( x ) =>• Q( x ) for all x in a dom ain S, we will often include the initial assum ption that P (x ) is true for an arbitrary elem ent x 6 S in order to solidify this technique in your m ind.
3 .3 P r o o f b y C o n tr a p o sitiv e
F or statem ents Pand Q,the c o n tra p o s itiv e o f the im plication P => Q isthe im plication (~<2) ( ' "P )- For exam ple, for P i : 3 is odd and Pi_ : 57 is prim e, the contrapositive o f the im plication
P\ => P2: I f 3 is odd, then 57 is prim e.
is the im plication
( ~ P 2) => ( ~ P i ) : I f 57 is n o t prim e, then 3 is even.
The m ost im portant feature o f the contrapositive ( ~ g ) =>• ( ~ / >) is that it is logically equivalent to P => Q.This fact is stated form ally as a theorem and is verified in the truth table show n in Figure 3.2.
Theorem 3.9 For every tw o statem ents P a n d Q, the im plication P => Q an d its contrapositive are logically equivalent; that is,
p
=> Q = (~G) => (~P).
Let
P ( x) : x = 2. and Q(x) : x 2 = 4.
w here x e R . T he contrapositive o f the im plication
P ( x) =^* Q(x) : I f x = 2, then x 2 = 4.
is the im plication
3.3 Proof by Contrapositive 85
P Q P =» Q ~ P ~ Q
T T T F F T
T F F F T F
F T T T F T
F F T T T T
Figure 3 .2 T h e l o g i c a l e q u i v a l e n c e o f a n i m p l i c a t i o n a n d its c o n t r a p o s i t i v e
=> C-' /^C.v>> : I f x 2 ^ 4, t h e n x ^ 2.
Suppose that we w ish to prove a result (or theorem ) w hich is expressed as
L et x 6 S. I f / J(x). then <2(x )- (3.2) or as
F or all x e S. if P( x ) , then Q( x ) . (3.3) We have seen that a p ro o f o f such a result consists o f establishing the truth o f the im plication P ( x ) =>■ Q( x ) for all x
e S.
If it can be show n that ( ~ ( X v ) ) ->■ ( ~ / J (x)) is true for all xe
S, then P ( x ) =$> Q( x ) is true for all x €S. A
p r o o f by c o n tra p o s itiv e o f the result (3.2) (or o f (3.3)) is a direct p ro o f o f its contrapositive:L et x €
S.
I f ~<2(x), then ~ P ( x ) . orF or all x €
S,
if ~£>(x), then ~ P ( x ) .Thus to give a p ro o f by contrapositive o f (3.2) (or o f (3.3)), we assum e that ~ g ( x ) is true for an arbitrary elem ent x
e S
and show that ~ /D(x) is true for this elem ent x.T here are certain types o f results w here a p ro o f by contrapositive is preferable or p erhaps even essential. We now give som e exam ples to illustrate this m ethod o f proof.
Result 3.10 L e t x e Z. I f 5x — 7 is even, then x is odd.
P roof A ssum e that x is even. T hen x = 2a for som e integer a. So
5x - 7 = 5 (2 a) - 7 = 10a - 7 = 10a - 8 + 1 = 2 (5a - 4) + 1.
S ince 5a — 4 e Z, the integer 5x — 7 is odd. g
PROOF ANALYSIS Som e com m ents are now in order. The goal o f R esult 3.10 w as to prove P ( x ) =>• Q( x ) for all x e Z, w here P ( x ) : 5x — 7 is even, and Q( x ) : x is odd. Since w e chose to give a p ro o f by contrapositive, we gave a direct p ro o f o f (~£>(x)) ( ~ P ( j J ) for all x e Z.
H ence the p ro o f began by assum ing that x is not odd; that is, x is even. The object then w as to show that 5x — 7 is odd.
I f w e had attem pted to prove R esult 3.10 w ith a direct proof, then w e w ould have begun by assum ing that 5x - 7 is even fo r an arbitrary integer x . T hen 5x - 7 = 2a for som e integer a. So x = (2a + 7 ) /5 . We then w ould w ant to show that x is odd. W ith the expression w e have for x , it is n ot even clear that x is an integer, m uch less that x is an o d d
Chapter 3 Dircci Proof' and Proof by Contrapositive
integer, although, o f course, we w ere told in the statem ent o f R esult 3.10 that the dom ain o f x is the set o f integers. T herefore, it is not only that a p ro o f by contrapositive provides us w ith a rath er sim ple m ethod o f proving R esult 3.10, it m ay not be im m ediately clear how or w hether a direct p ro o f can be used.
H ow did w e know beforehand that it is a p ro o f by contrapositive that w e should u se here? T his is n o t as difficult as it m ay appear. If w e use a direct proof, then we begin by assum ing that 5x — 7 is even for an arbitrary integer x ; w hile if we use a p ro o f by contrapositive, then we begin by assum ing that x is even. T herefore, using a pro o f by contrapositive allow s us to w ork w ith
x
initially rath er than the m ore com plicatedexpression 5* — 7. +
In all o f the exam ples that w e have seen so far, w e have considered only im plications.
N ow we look at a biconditional.
Result 3.11 L e t x e Z. Then 1 I x — 7 is even i f a n d only i f x is odd.
P roof T here are tw o im plications to prove here, nam ely,
(1) if x is odd, then 11* — 7 is even and (2) if 1 l x — 7 is even, then x is odd.
We begin w ith (1). In this case, a direct p ro o f is appropriate. A ssum e that x is odd. Then x = 2r + 1, w here r e Z. So
l l x - 7 = 11(2r + 1) - 7 = 22r + 1 1 - 7 = 22r + 4 = 2(11 r + 2).
Since H r + 2 is an integer, 1 lx — 7 is even.
We now prove (2), w hich is the converse o f (1). W e use a p ro o f by contrapositive here. A ssum e that x is even. T hen x = 2 s, w here s € Z . T herefore,
l l x - 7 = 11 (2s) — 7 = 22s - 7 = 22s - 8 + 1 = 2 ( 1 I s — 4 ) + 1.
Since 11s — 4 is an integer, l l x — 7 is odd. ■
A com m ent concerning the statem ents o f R esults 3.10 and 3.11 bears repeating here.
These results begin w ith the sentence: L e tx e Z. This, o f course, is inform ing us that the dom ain in this case is Z. That is, w e are being told that x represents an integer. We need n o t state this assum ption in the proof. T he sentence “L et x € Z.” is com m only called an
“overriding” assum ption or hypothesis and so x is assum ed to be an integer throughout the proofs o f R esults 3.10 and 3.11.
In the p ro o f o f R esu lt 3.11, w e discussed o ur p la n o f attack. N am ely, w e stated that there w ere tw o im plications to prove, and we specifically stated each. O rdinarily w e d o n ’t include such inform ation w ithin the p ro o f— unless the p roof is quite long, in w hich case a roadm ap indicating the steps w e plan to take m ay be helpful. We give an additional exam ple o f this type, w here this tim e a m ore conventional condensed p ro o f is presented. The follow ing exam ple w ill be useful to us in the future, thus we refer to it as a theorem .
Theorem 3.12 L e t x e Z. Then x 2 is even i f a n d only i f x is even.
3.3 Proof by Coiitrapositive 8 7
P roof A ssum e th a tM is even. T hen x = 2a for som e integer a. Therefore, x 2 = (2a)2 = 4 a 2 = 2(2 a 2).
B ecause 2 a 2 e Z, the integer x 2 is even.
F or the converse, assum e that x is odd. So x — 2b + I, w here b € Z. Then .r2 = (2b + l )2 = 4 b 2 + 4b + 1 = 2(2 Zr + 2b) + 1.
Since 2 b 2 + 2b is an integer, x 2 is odd. ■
Suppose now that you w ere asked to prove the follow ing result:
L et x £ Z. T hen x 2 is odd if and only if x is odd. (3.4) H ow w ould you do this? You m ight think o f proving the im plication “I f x is odd, then x 2 is odd.” by a direct p ro o f and its converse “I f x 2 is odd, then ,v is odd.” by a p ro o f by contrapositive, w here, o f course, the dom ain o f ,v is Z. I f we look at w hat is happening here, w e see that we are duplicating the p ro o f o f T heorem 3.12. This is no surprise w hatsoever. T heorem 3.12 states that if x is even, then x 2 is even; and if x 2 is even, then x is even. The contrapositive o f the first im plication is “If x 2 is odd, then x is odd,”
w hile the contrapositive o f the second im plication is “I f x is odd, then x 2 is odd.” In o ther w ords, (3.4) sim ply restates T heorem 3.12 in term s o f contrapositives. Thus (3.4) requires no p ro o f at all. It is essentially a restatem ent o f T heorem 3.12. A nd speaking o f restatem ents o f T heorem 3.12, w e need to recognize that this theorem can be restated in other w ays. F or exam ple, w e could restate
If x is an even integer, then x 2 is even.
as
The square o f every even integer is even.
H ence T heorem 3.12 could be stated as:
A n integer is even i f an d only i f its square is even.
It is not only useful to som etim es restate results in different m anners fo r variety, it is im portant to recognize w hat a result is saying regardless o f the m anner in w hich it m ay be stated.
A t this point, it is convenient to pause and discuss how theorem s (or results) can be used and w hy it is that w e m ay be interested in proving a p articular theorem . First, it is only by providing a p ro o f o f a theorem th at w e know for certain that the theorem is true and therefore have the rig h t to call it a theorem . A fundam ental reason w hy m athem aticians m ay w ant to give a p ro o f o f som e m athem atical statem ent is that they consider this a challenge— this is w hat m athem aticians do.
T his, in fact, brings up a question that m any m athem aticians consider o f greater im portance. W here do such statem ents com e from ? O f course, the answ er is that they com e from m athem aticians or students. H ow these people arrive at such questions does not follow any set rule. B ut this deals w ith the creative aspect o f m athem atics. Some
88 Chapter 3 Direct Proof am i Proof by Contrapositive that the m ethod o f p ro o f used could be applied to prove som ething even m ore interesting.
(W hat is interesting, o f course, is quite subjective.) M ore than likely however, a person has observed som e relationship that exists in an exam ple being considered that appears to occur in a m ore general setting. T his individual then attem pts to show that this is the case by giving a proof. This entire process involves the idea o f conjectures (guesses) and trying to show the accuracy o f a conjecture. W e 'll discuss this at greater length later. contrapositive o f this result, we m ay be h eaded for difficulties. T here is, however, another approach. E ven though we m ust be very careful about w hat we are assum ing, from w hat we know about even and odd integers, it appears that if 5x — 7 is odd, then x m ust be
3.4 Proof by Cases 8 9
P roof L et 5 x — 1 be an odd integer. By L em m a 3.13. the integer x is even. Since x is even, x 2z for som e integer z . Thus
9 x + 2 = 9(2z) + 2 = 18z + 2 = 2(9z + 1).
B ecause 9z + 1 is an integer, 9 x + 2 is even. *
So, w ith the aid o f L em m a 3.13, w e have produced a very u ncom plicated (and, hopefully, easy-to-follow ) p ro o f o f R esult 3.14.
The m ain reason for presenting R esult 3.14 w as to show how h elpful a lem m a can be in producing a p ro o f o f another result. H ow ever, having ju st said this, we now show how we can prove R esult 3.14 w ithout the aid o f a lem m a, by perform ing a b it o f algebraic m anipulation.
Alternative P roof A ssum e that 5x — 7 is odd. T hen 5x — 7 = 2n + 1 for som e integer n. O bserve that o f Result 3.14
9x + 2 = (5x - 7) + (4x + 9) = 2n + 1 + 4x + 9
— 2/7 -(- 4x -f- 10 = 2(n H- 2x -j- 5).
B ecause n + 2x + 5 is an integer, 9x + 2 is even. ■
You m ay p refer one p ro o f o f R esult 3.14 over the other. W hether you do or not, it is im portant to know that tw o different m ethods can be used. T hese m ethods m ight prove to be useful for future results you encounter. A lso, you m ight th ink w e used a trick to give the second p ro o f o f R esult 3.14; but, as we w ill see, if the sam e “tric k ” can be used often, then it becom es a technique.
Result 3.14 L e t x a Z. I f 5 x - 7 is o d d , then 9 x -f 2 is even.
3.4 P ro o f by Cases
W hile attem pting to give a p ro o f o f a m athem atical statem ent concerning an elem ent x in som e set 5, it is som etim es useful to observe that x possesses one o f tw o or m ore properties. A com m on property w hich x m ay possess is that o f belonging to a particular subset o f S. I f we can verify the truth o f the statem ent for each property that x m ay have, then w e have a p ro o f o f the statem ent. Such a p ro o f is then divided into parts called cases, one case for each property th at x m ay possess o r for each subset to w hich x m ay belong. This m ethod is called proof by cases. Indeed, it m ay be useful in a p ro o f by cases to further divide a case into other cases, called subcases.
F or exam ple, in a p ro o f o f V/? e Z. R( n) , it m ight be convenient to use a p ro o f by cases w hose p ro o f is divided into the tw o cases
Case 1. n is even, and C ase 2. n is odd.
O ther possible proofs by cases m ight involve proving Vx e R, P ( x ) using the cases C ase 1 .x = 0. C ase 2. x < 0 and Case 3. x > 0.
Chapter 3 Direct Proof and Proof by Contrapositive
Result 3.15
P roof
Theorem 3.16
P roof
A lso, we m ight attem pt to prove i n e N, P( n ) using the cases C ase 1. n = 1 and C ase 2. n > 2.
Furtherm ore, for S = Z — {0}, w e m ight try to prove Vx, y e S, P( x . -y) by using the cases:
C ase 1. x y > 0 and C ase 2. x y < 0.
C ase 1 could, in fact, be divided into tw o subcases:
Subcase 1. 1. x > 0 a n d y > 0. and Subcase 1.2. x < 0 a n d y < 0, w hile C ase 2 could be divided into the tw o subcases:
Subcase 2.1. x > 0 an d y < 0 . and Subcase 2.2. x < 0 a n d y > 0.
L e t’s look at an exam ple o f a p ro o f by cases.
I f n e Z, then n 2 + 3n + 5 is an odd integer.
We proceed by cases, according to w hether n is even or odd.
C ase 1. n is even. T hen n = 2x for som e x e Z. So
n 2 + 3n + 5 = (2 x )2 + 3(2 v) + 5 = 4 x 2 + 6x + 5 = 2 (2 x 2 + 3x + 2) + 1.
Since 2 x 2 + 3x + 2 6 Z, the integer n 2 + 3n + 5 is odd.
C ase 2. n is odd. T hen n = 2 y + 1, w here y e Z. Thus
n~ + 3n + 5 = (2_y + I )2 + 3(2y + 1) + 5 = 4 y 2 + lOy + 9 - 2 ( 2 y + 5y + 4) + 1.
B ecause 2 y 2 + 5y + 4 e Z, the integer n 2 + 3n + 5 is odd. ■
Two integers x and y are said to be o f th e sa m e p a r ity if x and y are both even or are both odd. The integers x and y are o f o p p o site p a r ity if one o f x and y is even and the other is odd. F or exam ple, 5 and 13 are o f the sam e parity, w hile 8 and 11 are o f o pposite parity. B ecause the definition o f tw o integers having the sam e (or opposite) parity requires the tw o integers to satisfy one o f tw o properties, any result containing these term s is likely to be proved by cases. T he follow ing theorem presents a characterization o f tw o integers that are o f the sam e parity.
L e t x , y e Z. T hen x a n d y are o f the sam e p a rity i f a n d only i f x + y is even.
F irst, assum e that x and y are o f the sam e parity. We consider tw o cases.
C ase 1. x a n d y are even. T hen x = 2a and v = 2b for som e integers a and b. So x + y = 2a + 2b = 2(a + b). Since a + b e Z, the integer x + y is even.
C ase 2. x an d y are odd. T hen x = 2a + I and y = 2b + 1, w here a , b e Z. T herefore, x + y = (2 a + 1) + (2b + I) — 2a + 2b + 2 = 2 (a + b + 1).
Since a + b + 1 is an integer, x + y is even.
3.4 Proof by Cases 91
PROOF ANALYSIS
Theorem to Prove PROOFSTRATEGY
For the converse, assum e that x and y are o f opposite parity. A gain, w e consider tw o cases.
C ase 1. x is even an d y is odd. Then x = 2a and y = 2b + 1, w here a, b € Z. T hen x + y — 2a + (2b + 1) = 2 (a + b) + 1.
Since a + b e Z, the integer x + y is odd.
C ase 2. x is od d an d y is even. The p ro o f is sim ilar to the p ro o f o f the preceding case
and is therefore om itted. ■
A com m ent concerning the p ro o f o f T heorem 3.16 is useful here. A lthough there is alw ays som e concern w hen om itting steps or proofs, it should be clear that it is truly a w aste o f effort by w riter and reader alike to give a p ro o f o f the case w hen x is odd and v is even in T heorem 3.16. Indeed, there is an alternative w hen the converse is considered.
F or the converse, assum e that x and y are o f opposite parity. W ithout loss o f gener
ality, assum e that x is even and y is odd. T hen x = 2a and y = 2b + 1, w here a , b e Z.
T hen
x + y = 2a + (2b + 1) = 2(a + b) + 1.
Since a + b e Z, the integer x + y is odd. ♦
We used the phrase w ith o u t loss o f g e n e ra lity (som e abbreviate this as W O LO G or W L O G ) to indicate that the proofs o f the tw o situations are sim ilar, so the p ro o f o f only one o f these is needed. S om etim es it is rather subjective to say that tw o situations are similar. We present one additional exam ple to illustrate this.
L et a and b be integers. T hen ab is even if and only if a is even or b is even.
B efore w e begin a p ro o f o f this result (Theorem 3.17 below ), le t’s see w hat w e w ill be required to show. We need to prove tw o im plications, nam ely, (1) If a is even or b is even, then a b is even and (2) if a b is even, then a is even o r b is even. We consider (1) first. A direct p ro o f seem s appropriate. H ere, we w ill assum e that a is even or b is even. W e could give a p ro o f by cases: (i) a is even, (ii) b is even. O n the other hand, since the proofs o f these cases will certainly be sim ilar, we could say, w ithout loss o f generality, that a is even. We w ill see that it is unnecessary to m ake any assum ption about b.
If w e w ere to give a direct p ro o f o f (2), then we w ould begin by assum ing that a b is even, say ab = 2 k for som e integer k. B ut how could we deduce any inform ation about a and b individually? L e t’s try another approach. If w e use a p ro o f by contrapositive, then we w ould begin by assum ing that it is not the case that a is even or b is even. This is exactly the situation covered by one o f D e M o rg an ’s laws:
~ c p V Q ) is logically equivalent to (~ .P ) A (~<2).
It is im portant not to forget this. In this case, w e have P : a is even, and Q : b is even.
So the negation o f “ a is even or b is even” is “a is odd a n d b is odd.” ♦
L e t’s now prove this result.
92 Chapter 3 Direct Proof and Proof by Contrapositive
Theorem 3.17 L e t a an d b be integers. Then ab is even i f a n d only i f a is even or b is even.
P roof F irst, assum e that a is even or b is even. W ithout loss o f generality, let a be even. T hen a = 2 x for som e integer x . Thus a b = (.2 x )b = 2(xb). Since x b is an integer, a b is even.
For the converse, assum e that a is odd and b is odd. T hen a — 2 x + 1 and b = 2y + 1, w here x , y e Z. H ence
ab = ( 2x + l ) ( 2 y + 1) = 4 x y + 2x + 2 y + 1 = 2 {2 xy + x + y ) + 1.
Since 2 x y + x + y is an integer, a b is odd. m
3.5 P ro o f E valuation s
We have now stated several results and have given a p ro o f o f each result (som etim es preceding a p ro o f by a p ro o f strategy or follow ing the p ro o f w ith a p ro o f analysis). L e t’s reverse this process by giving an exam ple o f a p ro o f o f a result but not stating the result being proved. We w ill follow the p ro o f w ith several options for the statem ents o f the result being proved.
Exam ple 3.18 G iven below is a p ro o f o f a result.
P roof A ssum e that n is an odd integer. T hen n = 2 k + \ for som e integer k. Then 3n — 5 = 3(2k + 1) - 5 = 6* + 3 - 5 = 6k - 2 = 2(3k - 1).
Since 3k — 1 is an integer, 3n — 5 is even. ■
W hich o f the follow ing is proved above?
(1) 3n — 5 is an even integer.
(2) If n is an odd integer, then 3n — 5 is an even integer.
(3) Let n be an integer. I f 3n — 5 is an even integer, then n is an odd integer.
(4) Let n be an integer. If 3n — 5 is an odd integer, then n is an even integer.
The correct answ ers are (2) and (4). The p ro o f given is a direct p ro o f o f (2) and a p ro o f by contrapositive o f (4). T he sentence ( I) is an open sentence, not a statem ent, and is only the conclusion o f (2). S tatem ent (3) is the converse o f (2). $
W hen learning any m athem atical subject, it is not the least bit unusual to m ake
W hen learning any m athem atical subject, it is not the least bit unusual to m ake