Katedra matematiky - Katolícka univerzita v Ružomberku
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Katedra matematiky - Katolícka univerzita v Ružomberku

Katedra matematiky - Katolícka univerzita v Ružomberku Katedra matematiky - Katolícka univerzita v Ružomberku

math.ku.sk
12.07.2015 Views

64 Jan Kopka, Leonard Frobisher, George Feissner3 = 1·3 = 0+1+25 = 1·5 = −1+0+1+2+37 = 1·7 = −2+(−1)+0+1+2+3+410 = 2·5 = 0+1+2+3+414 = 2·7 = −1+0+1+2+3+4+5.However, this last experiment has shown us exactly what we need to do toshow that Conjecture 3a is true. We present it as a proof.Proof of conjecture 3a): Let positive integer number n have an odd factor2k +1. Then n can be written in the form (1). Now we know, that n canbe written also in the formn = (g−k)+···+(g−2)+(g−1)+g+(g+1)+(g+2)+···+(g+k) (2)where g is also factor of n.How can we get only two or more positive numbers on the right hand sideof formula (2)? If there is zero, we delete it. All we have to do is counteractthe negative terms by the corresponding positive terms. There are morepositive terms than negative ones because the total sum is positive. Afterthis procedure we get the even number of consecutive positive terms.So after this proof we rename conjecture 3a as a theorem:Theorem 1: If a positive integer n has an odd factor other than 1, then itcan be written as the sum of consecutive positive integers.Indeed, our investigation has given us more than what the theorem states;not only do we know that it can be done, but we know how to do it!Now we have enough experiences to believe that the opposite theorem totheorem 1 also holds.Theorem 2: If a positive integer n can be written as the sum of consecutivepositive integers, then it must have an odd factor other than 1.Proof: We writenasn = g 1 +g 2 +···+g r whereg 1 ,g 2 ,...g r are consecutivepositive integers. We consider two cases:(a) r is an odd number(b) r is an even number.Case (a): We can write n in the form shown in Formula 1, and then g willbe the number in the middle of the sum. The fact that r is odd guaranteesthere is a middle term. Then, exactly as in our experimentation, we canthen write n in the form n = g? (2k + 1) so n has an odd factor greaterthan 1, namely 2k +1.

Conjecturing: An Investigation Involving Positive Integers 65Case (b): We are unable to proceed as in Case (a) since now there is nomiddle term. However, we "augment" the sum g 1 +g 2 +···+g r by addingto it−(g 1 −1)+···+(−2)+(−1)+0+1+2+···+(g 1 −1).Note first that we have really just added on 0, since all the positive termscancel with the corresponding negative ones. Thus, the new sum is stillequal to n. Second, we have added on an odd number of terms - an equalnumber of positive and negative terms and 0. Since we began with an evennumber of terms in the sum g 1 +g 2 +···+g r we now have an odd number ofterms in the augmented sum, namely an odd number of consecutive integerswhose sum is n. The situation is now exactly as it was in Case (a), so wecan conclude that n has an odd factor greater than 1.It is difficult to say exactly how this "trick"of augmenting the sum in Case(b) was conceived. That is part of the creative side of mathematics. However,we can say with confidence that had we not been thinking of sumsof consecutive integers and of Formula (2), the trick would not have beenthought of.We now put both theorems together:Theorem 3: A positive integer n can be written as the sum of consecutivepositive integers if and only if the positive integer n has an odd factor otherthan one.Corollary of theorem 3 is conjecture 2. The powers of two have no odd factorsexcept 1, so these powers cannot be expressed as the sum of consecutivepositive integers. Therefore we can rename conjecture 2 as theorem 4:Theorem 4: Powers of two are not the sum of consecutive positive integers.Now we are inspired for another investigation.Some positive integers can be written as a sum of consecutive positive integersin different ways. For example:21 = 3·7 21 = (0+)1+2+3+4+5+6= 6+7+8= 10+1130 = 2·3·5 30 = 9+10+11= 4+5+6+7+8= 6+7+8+9We pose a new problem.

Conjecturing: An Investigation Involving Positive Integers 65Case (b): We are unable to proceed as in Case (a) since now there is nomiddle term. However, we "augment" the sum g 1 +g 2 +···+g r by addingto it−(g 1 −1)+···+(−2)+(−1)+0+1+2+···+(g 1 −1).Note first that we have really just added on 0, since all the positive termscancel with the corresponding negative ones. Thus, the new sum is stillequal to n. Second, we have added on an odd number of terms - an equalnumber of positive and negative terms and 0. Since we began with an evennumber of terms in the sum g 1 +g 2 +···+g r we now have an odd number ofterms in the augmented sum, namely an odd number of consecutive integerswhose sum is n. The situation is now exactly as it was in Case (a), so wecan conclude that n has an odd factor greater than 1.It is difficult to say exactly how this "trick"of augmenting the sum in Case(b) was conceived. That is part of the creative side of mathematics. However,we can say with confidence that had we not been thinking of sumsof consecutive integers and of Formula (2), the trick would not have beenthought of.We now put both theorems together:Theorem 3: A positive integer n can be written as the sum of consecutivepositive integers if and only if the positive integer n has an odd factor otherthan one.Corollary of theorem 3 is conjecture 2. The powers of two have no odd factorsexcept 1, so these powers cannot be expressed as the sum of consecutivepositive integers. Therefore we can rename conjecture 2 as theorem 4:Theorem 4: Powers of two are not the sum of consecutive positive integers.Now we are inspired for another investigation.Some positive integers can be written as a sum of consecutive positive integersin different ways. For example:21 = 3·7 21 = (0+)1+2+3+4+5+6= 6+7+8= 10+1130 = 2·3·5 30 = 9+10+11= 4+5+6+7+8= 6+7+8+9We pose a new problem.

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