# Lacsap’s fractions

Lacsap’s Fractions IB Math 20 Portfolio By: Lorenzo Ravani Lacsap’s Fractions Lacsap is backward for Pascal. If we use Pascal’s triangle we can identify patterns in Lacsap’s fractions. The goal of this portfolio is to ? nd an equation that describes the pattern presented in Lacsap’s fraction. This equation must determine the numerator and the denominator for every row possible. Numerator Elements of the Pascal’s triangle form multiple horizontal rows (n) and diagonal rows (r). The elements of the ? rst diagonal row (r = 1) are a linear function of the row number n. For every other row, each element is a parabolic function of n.

Where r represents the element number and n represents the row number. The row numbers that represents the same sets of numbers as the numerators in Lacsap’s triangle, are the second row (r = 2) and the seventh row (r = 7). These rows are respectively the third element in the triangle, and equal to each other because the triangle is symmetrical. In this portfolio we will formulate an equation for only these two rows to ? nd Lacsap’s pattern. The equation for the numerator of the second and seventh row can be represented by the equation: (1/2)n * (n+1) = Nn (r) When n represents the row number.

And Nn(r) represents the numerator Therefore the numerator of the sixth row is Nn(r) = (1/2)n * (n+1) Nn(r) = (1/2)6 * (6+1) Nn(r) = (3) * (7) Nn(r) = 21 Figure 2: Lacsap’s fractions. The numbers that are underlined are the numerators. Which are the same as the elements in the second and seventh row of Pascal’s triangle. Figure 1: Pascal’s triangle. The circled sets of numbers are the same as the numerators in Lacsap’s fractions. Graphical Representation The plot of the pattern represents the relationship between numerator and row number. The graph goes up to the ninth row.

The rows are represented on the x-axis, and the numerator on the y-axis. The plot forms a parabolic curve, representing an exponential increase of the numerator compared to the row number. Let Nn be the numerator of the interior fraction of the nth row. The graph takes the shape of a parabola. The graph is parabolical and the equation is in the form: Nn = an2 + bn + c The parabola passes through the points (0, 0) (1, 1) and (5, 15) At (0, 0): 0 = 0 + 0 + c ! ! At (1, 1): 1 = a + b ! ! ! At (5, 15): 15 = 25a + 5b ! ! ! 15 = 25a + 5(1 – a) ! 15 = 25a + 5 – 5a ! 15 = 20a + 5 ! 10 = 20a! ! ! ! ! ! ! therefore c = 0 therefore b = 1 – a Check with other row numbers At (2, 3): 3 = (1/2)n * (n+1) ! (1/2)(2) * (2+1) ! (1) * (3) ! N3 = (3) therefore a = (1/2) Hence b = (1/2) as well The equation for this graph therefore is Nn = (1/2)n2 + (1/2)n ! which simpli? es into ! Nn = (1/2)n * (n+1) Denominator The difference between the numerator and the denominator of the same fraction will be the difference between the denominator of the current fraction and the previous fraction. Ex. If you take (6/4) the difference is 2. Therefore the difference between the previous denominator of (3/2) and (6/4) is 2. ! Figure 3: Lacsap’s fractions showing differences between denominators Therefore the general statement for ? nding the denominator of the (r+1)th element in the nth row is: Dn (r) = (1/2)n * (n+1) – r ( n – r ) Where n represents the row number, r represents the the element number and Dn (r) represents the denominator. Let us use the formula we have obtained to ? nd the interior fractions in the 6th row. Finding the 6th row – First denominator ! ! ! ! ! ! ! ! ! ! ! ! – Second denominator ! ! ! ! ! ! ! ! ! ! ! ! ! denominator = 6 ( 6/2 + 1/2 ) – 1 ( 6 – 1 ) ! = 6 ( 3. 5 ) – 1 ( 5 ) ! 21 – 5 = 16 denominator = 6 ( 6/2 + 1/2 ) – 2 ( 6 – 2 ) ! = 6 ( 3. 5 ) – 2 ( 4 ) ! = 21 – 8 = 13 ! ! -Third denominator ! ! ! ! ! ! ! ! ! ! ! ! – Fourth denominator ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! – Fifth denominator ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! denominator = 6 ( 6/2 + 1/2 ) – 3 ( 6 – 3 ) ! = 6 ( 3. 5 ) – 3 ( 3 ) ! = 21 – 9 = 12 denominator = 6 ( 6/2 + 1/2 ) – 2 ( 6 – 2 ) ! = 6 ( 3. 5 ) – 2 ( 4 ) ! = 21 – 8 = 13 denominator = 6 ( 6/2 + 1/2 ) – 1 ( 6 – 1 ) ! = 6 ( 3. 5 ) – 1 ( 5 ) ! = 21 – 5 = 16 ! ! We already know from the previous investigation that the numerator is 21 for all interior fractions of the sixth row.

Using these patterns, the elements of the 6th row are 1! (21/16)! (21/13)! (21/12)! (21/13)! (21/16)! 1 Finding the 7th row – First denominator ! ! ! ! ! ! ! ! ! ! ! ! – Second denominator ! ! ! ! ! ! ! ! ! ! ! ! – Third denominator ! ! ! ! ! ! ! ! ! ! ! ! – Fourth denominator ! ! ! ! ! ! ! ! ! ! ! ! ! ! denominator = 7 ( 7/2 + 1/2 ) – 1 ( 7 – 1 ) ! = 7(4)–1(6) ! = 28 – 6 = 22 denominator = 7 ( 7/2 + 1/2 ) – 2 ( 7 – 2 ) ! = 7(4)–2(5) ! = 28 – 10 = 18 denominator = 7 ( 7/2 + 1/2 ) – 3 ( 7 – 3 ) ! = 7(4)–3(4) ! = 28 – 12 = 16 denominator = 7 ( 7/2 + 1/2 ) – 4 ( 7 – 3 ) ! = 7(4)–3(4) ! = 28 – 12 = 16 ! ! ! ! ! ! Fifth denominator ! ! ! ! ! ! ! ! ! ! ! ! – Sixth denominator ! ! ! ! ! ! ! ! ! ! ! ! denominator = 7 ( 7/2 + 1/2 ) – 2 ( 7 – 2 ) ! ! = 7(4)–2(5) ! ! = 28 – 10 = 18 ! ! denominator = 7 ( 7/2 + 1/2 ) – 1 ( 7 – 1 ) ! = 7(4)–1(6) ! = 28 – 6 = 22 We already know from the previous investigation that the numerator is 28 for all interior fractions of the seventh row. Using these patterns, the elements of the 7th row are 1 (28/22) (28/18) (28/16) (28/16) (28/18) (28/22) 1 General Statement To ? nd a general statement we combined the two equations needed to ? nd the numerator and to ? nd the denominator. Which are (1/2)n * (n+1) to ? d the numerator and (1/2)n * (n+1) – n( r – n) to ? nd the denominator. By letting En(r) be the ( r + 1 )th element in the nth row, the general statement is: En(r) = {[ (1/2)n * (n+1) ] / [ (1/2)n * (n+1) – r( n – r) ]} Where n represents the row number and r represents the the element number. Limitations The ‘ 1’ at the beginning and end of each row is taken out before making calculations. Therefore the second element in each equation is now regarded as the ? rst element. Secondly, the r in the general statement should be greater than 0. Thirdly the very ? rst row of the given pattern is counted as the 1st row.

Lacsap’s triangle is symmetrical like Pascal’s, therefore the elements on the left side of the line of symmetry are the same as the elements on the right side of the line of symmetry, as shown in Figure 4. Fourthly, we only formulated equations based on the second and the seventh rows in Pascal’s triangle. These rows are the only ones that have the same pattern as Lacsap’s fractions. Every other row creates either a linear equation or a different parabolic equation which doesn’t match Lacsap’s pattern. Lastly, all fractions should be kept when reduced; provided that no fractions common to the numerator and the denominator are to be cancelled. ex. 6/4 cannot be reduced to 3/2 ) Figure 4: The triangle has the same fractions on both sides. The only fractions that occur only once are the ones crossed by this line of symmetry. 1 Validity With this statement you can ? nd any fraction is Lacsap’s pattern and to prove this I will use this equation to ? nd the elements of the 9th row. The subscript represents the 9th row, and the number in parentheses represents the element number. – E9(1)!! ! – First element! ! ! ! ! ! ! ! ! ! ! ! ! – E9(2)!! ! – Second element! ! ! ! ! ! ! ! ! ! ! ! ! – E9(3)!! ! – Third element! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! {[ n( n/2 + 1/2 ) ] / [ n( n/2 + 1/2 ) – r( n – r) ]} {[ 9( 9/2 + 1/2 ) ] / [ 9( 9/2 + 1/2 ) – 1( 9 – 1) ]} {[ 9( 5 ) ] / [ 9( 5 ) – 1( 8 ) ]} {[ 45 ] / [ 45 – 8 ]} {[ 45 ] / [ 37 ]} 45/37 {[ n( n/2 + 1/2 ) ] / [ n( n/2 + 1/2 ) – r( n – r) ]} {[ 9( 9/2 + 1/2 ) ] / [ 9( 9/2 + 1/2 ) – 2( 9 – 2) ]} {[ 9( 5 ) ] / [ 9( 5 ) – 2 ( 7 ) ]} {[ 45 ] / [ 45 – 14 ]} {[ 45 ] / [ 31 ]} 45/31 {[ n( n/2 + 1/2 ) ] / [ n( n/2 + 1/2 ) – r( n – r) ]} {[ 9( 9/2 + 1/2 ) ] / [ 9( 9/2 + 1/2 ) – 3 ( 9 – 3) ]} {[ 9( 5 ) ] / [ 9( 5 ) – 3( 6 ) ]} {[ 45 ] / [ 45 – 18 ]} {[ 45 ] / [ 27 ]} 45/27 E9(4)!! ! – Fourth element! ! ! ! ! ! ! ! ! ! ! ! ! – E9(4)!! ! – Fifth element! ! ! ! ! ! ! ! ! ! ! ! ! – E9(3)!! ! – Sixth element! ! ! ! ! ! ! ! ! ! ! ! ! – E9(2)!! ! – Seventh element! ! ! ! ! ! ! ! ! ! ! ! ! – E9(1)!! ! – Eighth element! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! [ n( n/2 + 1/2 ) ] / [ n( n/2 + 1/2 ) – r( n – r) ]} {[ 9( 9/2 + 1/2 ) ] / [ 9( 9/2 + 1/2 ) – 4( 9 – 4) ]} {[ 9( 5 ) ] / [ 9( 5 ) – 4( 5 ) ]} {[ 45 ] / [ 45 – 20 ]} {[ 45 ] / [ 25 ]} 45/25 {[ n( n/2 + 1/2 ) ] / [ n( n/2 + 1/2 ) – r( n – r) ]} {[ 9( 9/2 + 1/2 ) ] / [ 9( 9/2 + 1/2 ) – 4( 9 – 4) ]} {[ 9( 5 ) ] / [ 9( 5 ) – 4( 5 ) ]} {[ 45 ] / [ 45 – 20 ]} {[ 45 ] / [ 25 ]} 45/25 {[ n( n/2 + 1/2 ) ] / [ n( n/2 + 1/2 ) – r( n – r) ]} {[ 9( 9/2 + 1/2 ) ] / [ 9( 9/2 + 1/2 ) – 3 ( 9 – 3) ]} {[ 9( 5 ) ] / [ 9( 5 ) – 3( 6 ) ]} {[ 45 ] / [ 45 – 18 ]} {[ 45 ] / [ 27 ]} 45/27 {[ n( n/2 + 1/2 ) ] / [ n( n/2 + 1/2 ) – r( n – r) ]} {[ 9( 9/2 + 1/2 ) ] / [ 9( 9/2 + 1/2 ) – 2( 9 – 2) ]} {[ 9( 5 ) ] / [ 9( 5 ) – 2 ( 7 ) ]} {[ 45 ] / [ 45 – 14 ]} {[ 45 ] / [ 31 ]} 45/31 {[ n( n/2 + 1/2 ) ] / [ n( n/2 + 1/2 ) – r( n – r) ]} {[ 9( 9/2 + 1/2 ) ] / [ 9( 9/2 + 1/2 ) – 1( 9 – 1) ]} {[ 9( 5 ) ] / [ 9( 5 ) – 1( 8 ) ]} {[ 45 ] / [ 45 – 8 ]} {[ 45 ] / [ 37 ]} 45/37 From these calculations, derived from the general statement the 9th row is 1 (45/37)! ! (45/31)! ! (45/27)! (45/25)! (45/25)! (45/27) (45/31)! (45/37)! ! 1 Using the general statement we have obtained from Pascal’s triangle, and following the limitations stated, we will be able to produce the elements of any given row in Lacsap’s pattern. This equation determines the numerator and the denominator for every row possible.