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The identity 3*5 = 15 in different contextsby Walter VanniniBase 2 and an algebraic identity Base 4 and and difference of squares Base 2 and another algebraic identity Triangular Numbers Fibonacci numbers Sums of Fibonacci squares Three Times Generalized Fibonacci The Golden Ratio Mersenne primes Product of Sum of Squares Complex Numbers Quaternions An infinite product Concluding words Feedback IntroductionIt's often been said that all of mathematics is connected. This essay is about some of the mathematics that is very closely connected to the result "3*5=15". Amongst other things, that includes various algebraic identities, Fibonacci numbers, the Golden Ratio, Mersenne primes, sums of squares, complex numbers, and quaternions. Base 2 and an algebraic identityBy viewing 3*5=15 in base 2, it's 11*101=1111. This tells us that the identity can be viewed as (2+1)(4+1)=8+4+2+1, which is the identity resulting from substituting x=2 in the algebraic identity Base 4 and difference of squaresWriting the identity in base 4, it's 3*11 = 33. In base 4, 3 is not just 1+1+1, but it is also the largest one digit number. It corresponds to 9 in base 10, so that the identity is really analogous to 9*11= 99. The general equation is Base 2 and another algebraic identityWriting 3*5=15 in base 2 as 11*101=1111, the fact that 1 is the largest digit in base 2 means that there are analogues in base 10 where 1 is replaced by 9. They are: 99*101 = 9999 and 11*909=9999 (I'm going to ignore 11*901=9911, 19*101=1199 etc) both of which are really 9 * 11 * 101 = 9999 The underlying algebraic identity is Triangular Numbers15 is the triangular number (1+2+3+4+5). 3*5 = 15 can be viewed as the result of substituting n=5 in the the formula (n+1)/2 * n = (1+2+3+…+n) Fibonacci numbersRecall that the Fibonacci sequence has terms 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, … and can be defined recursively via: F_{n} = F_{n1} + F_{n2}F_{0} = 0 F_{1} = 1 The fact that 3 and 5 are in the sequence suggests that 3*5 = 15 is a special case of an identity involving Fibonacci numbers. Sums of Fibonacci squaresIt turns out that the product of the n^{th} Fibonacci number with the following Fibonacci number is the sum of the squares of the first n Fibonacci numbers. ie. F_{n} * F_{n+1} = F_{1}^{2} + F_{2}^{2} + … + F_{n}^{2}. Substituting the value n=4 in the above identity, we get F_{4} * F_{5} = F_{1}^{2} + F_{2}^{2} + F_{3}^{2} + F_{4}^{2}. 3 * 5 = 1^{2} + 1^{2} + 2^{2} + 3^{2}. Three Times Generalized FibonacciFor a Generalized Fibonacci sequence (ie a sequence satisfying the Fibonacci recurrence) we have 3 * G_{n} = G_{n+2}+G_{n2}The proof is straightforward:
Letting G_{n} be the Fibonacci sequence, and substituting n=5, we have 3 * F_{5} = F_{7} + F_{3} 3 * 5 = 13 + 2. The Golden RatioRecall that the golden ratio φ is (1+√5) ⁄ 2. An exact formula for the Fibonacci numbers is given by This instantly leads to the approximation In fact, F_{n} is always the closest integer to φ^{n} ⁄ √5.
Since 3 and 5 are the fourth and fifth Fibonacci numbers, we have that 3 is approximately φ^{4} ⁄ √5, and 5 is approximately φ^{5} ⁄ √5. This means that 3*5 must be approximately φ^{9} ⁄ 5.
So, the multiplicative identity 3*5=15 can be viewed as an integral approximation to the exact additive identity 4+5=9. Mersenne primes15 is 2^{2*2}1. There is an algebraic identity for n^{ab}  1, namely n^{ab}  1 = (n^{a}1)( 1+n^{a}+…+n^{aba} ).
The easiest way to see this is to realize that it's just Substituting n=2, a=2, b=2, we get 2^{2*2}=(2^{2}1)(1+2^{2}) ie 15=(3)(5). The above algebraic identity shows us that n^{m}1 cannot be prime if m is composite (given n ≥ 2 of course). 15 = 3*5 is the smallest example of the general fact "if q is composite, then 2^{q}1 is composite". The above result about composite numbers is equivalent to the result "if 2^{p}1 is a prime, then p is also a prime". Numbers of the form 2^{p}1 that are prime have been heavily studied, and have been given the name "Mersenne primes". Examples are 3 (= 2^{2}1) 7 (= 2^{3}1) and 31 (= 2^{5}1). 15 = 2^{4}1 can't be an example, since 4 is composite. Incidentally, p being prime doesn't necessarily imply that 2^{p}1 is prime. Consider 2^{11}1 = 2047 = 23*89. The largest known prime (as of August 2007) is the Mersenne prime 2^{32,582,657}1 Product of Sum of SquaresFibonacci showed that the product of the sum of two squares is always the sum of two squares. He did it by discovering the identity (a^{2} + b^{2}) (c^{2} + d^{2}) = (acbd)^{2} + (ad+bc)^{2} Euler showed that the product of the sum of four squares is always the sum of four squares. He did it by discovering the identity (a^{2} + b^{2} + c^{2} + d^{2}) (e^{2} + f^{2} + g^{2} + h^{2}) = (aebfcgdh)^{2} + (af+be+chdg)^{2} + (agbh+ce+df)^{2} + (ah+bgcf+de)^{2} But, the product of the sum of three squares is NOT always the sum of three squares. It turns out that the smallest example that demonstrates this fact about the sum of three squares is 3*5 =15
3 is the sum of three squares ( 1^{2} + 1 ^{2} + 1^{2} )
Complex NumbersFibonacci's identity is really a fact about multiplication of complex numbers. The norm of a complex number z = a + bi is defined to be a^{2}+b^{2} (equivalently (a+bi)*(abi)). Writing z_{1} = a+bi and z_{2} = c+di, Fibonacci's identity is the statement QuaternionsThe quaternions are a four dimensional generalization of the complex numbers. For the quaternion a+bi+cj+dk, define the norm to be a^{2}+b^{2}+c^{2}+d^{2} (equivalently (a+bi+cj+dk)*(abicjdk)). Writing q_{1} = a+bi+cj+dk and q_{2} = e+fi+gj+hk, Euler's identity is the statement If there were a three dimensional generalization of the complex numbers, then the corresponding norm construction would give us an identity for the product of the sums of three squares. The result "3*5=15" can be viewed as an obstruction to the existence of a three dimensional generalization of the complex numbers. To be concrete, if we were to consider some generalization of the complex numbers that involved numbers of the form a+bi+cj, then 3*5=15 would tell us that there's going to be a problem with the product (1+i+j)*(2+i). Of course, in the four dimensional world of quaternions, we have Applying the identity N(q_{1}) N(q_{2}) = N (q_{1}q_{2}), this becomes An infinite productSo far, we've seen the identities: x^{2}1 = (x1)(x+1) (from the base 4 interpretation) x^{4}  1 = (x1) (x+1) (x^{2}+1) (from a base 2 interpretation) This pattern can be continued, to yield further identities: x^{8}  1 = (x1) (x+1) (x^{2}+1) (x^{4}+1) x^{16}  1 = (x1) (x+1) (x^{2}+1) (x^{4}+1)(x^{8}+1)
The x=10 versions are: These algebraic identities can be rewritten as the sequence of identities: 1  x = (1x) 1  x^{2} = (1x) (1+x) 1  x^{4} = (1x) (1+x) (1+x^{2}) 1  x^{8} = (1x) (1+x) (1+x^{2}) (1+x^{4}) 1  x^{16} = (1x) (1+x) (1+x^{2}) (1+x^{4})(1+x^{8}) 1  x^{32} = (1x) (1+x) (1+x^{2}) (1+x^{4})(1+x^{8})(1+x^{16}) … Taking the limit makes sense for small x (more precisely, for x with absolute value less than 1) and this leads to the infinite product: 1 = (1x) (1+x) (1+x^{2}) (1+x^{4})(1+x^{8})(1+x^{16})… This tells us that: 1/1x = (1+x) (1+x^{2}) (1+x^{4})(1+x^{8})(1+x^{16})… Multiplying out the infinite product, this becomes the familiar 1/1x = 1 + x + x^{2} + x^{3} + x^{4} + x^{5} + … Concluding wordsI expect that there are many more mathematical results that can be somehow related to 3*5=15. Feel free to let me know about them. FeedbackIf you have corrections, additions, modifications, etc please let me know mailto:walterv@gbbservices.com
