www.inquirymaths.co.uk © 2005 Andrew Blair
Changing the conditions
The sum of two fractions equals their product.
If d > c, a is negative; and if (c - d) is not a factor of bc, a is fractional. Both results contradict the definition of a, b, c and d.
Mathematical notes
.............................................................................................
a, b, c and d are positive integers
In the two examples above, the
greater than one.
fractions
a a c c + = x b d b d
a c and in their simplest d b
form give a = c, i.e.
ad bc ac + = bd bd bd
5 5 7 , and, 2 3 4
7 . Is this always the case? 3
Therefore, ad + bc = ac bc = ac – ad
No, because when b = d, a ≠ c
bc = a(c – d) a =
Examples
bc cd
(1) If b = d = 2, and c = 6, then a =
Conditions c > d; and (c - d) is a factor 3 gives:
of bc.
3 6 3 6 + = x 2 2 2 2
(2) If b = d = 3 and c = 12, then a = Using the conditions and with c as the 4 gives:
independent variable, a, b, and d
4 12 4 12 + = x 3 3 3 3
(dependent variables) can follow: Conjectures Examples The following conjectures have arisen
(1) c = 5, d = 3, b = 4 a = 10 gives:
in Year 11 Higher tier GCSE lessons
10 5 10 5 + = x 4 3 4 3
from numerical examples. The attempts presented here to prove the
(2) c = 7, d = 3, b = 8, a = 14 gives:
conjectures true have come from
14 7 14 7 + = x 8 3 8 3
small groups of students.
1
www.inquirymaths.co.uk © 2005 Andrew Blair
(1) If the numerator is double the
(3) Alternative conjecture based on
denominator in both fractions, then the
number (2) above:
sum and product equal 4.
If the numerators are equal and odd, then the denominators are half of the numerator ±½.
2a 2b 2ab 2ab 4ab + = = =4 a b ab ab
To prove: and
2a 2b 4ab . = =4 a b ab
2n 1 2n 1 2n 1 2n 1 . = + n n 1 n n 1
.............................................................................................
(2) If the denominators are consecutive numbers, then the numerators are
Product:
equal and odd. Sum: Examples 5 5 25 + = 2 3 6
and
49 7 7 + = 3 4 12
To prove a = b when n and n + 1 are
2n 1 2n 1 (2n 1) 2 . = n(n 1) n n 1 2n 1 2n 1 + n n 1
=
(2n 1)(n 1) n(2n 1) n(n 1)
=
(2n 1)(n 1 n) n(n 1)
(2n 1) 2 = n(n 1)
denominators: a b a b + = . n n 1 n n 1
Therefore, a(n + 1) + bn = ab
n 1 n + =1 b a If n = 2, a = b = 5, confirms the conjecture. However, a = 4, b = 6 and n = 2 also satisfy the formula. Therefore 2 3 and represent a counter-example 4 6
and proves the conjecture false.
.............................................................................................
2
Q. E. D.
