"Result the same"

sophia

when some numbers are subtracted or divided the result will be the same . can you give the numbers ? (Fractional numbers can be used)

the soln is given as (4.5 , 3), (5 1/3, 4), (6 1/4, 5)

NB:- 5 1/3, should be read as 5 and 1 by 3
6 1/4,should be read as 6 and 1 by 4


richard heathfield

I found 37 solutions in the range (-10, -10) to (+10, +10), of which 20
were negative.

Qurqirish dragon

From the given examples, I hope you can see that there is an infinite family of solutions of the form:
x*(1 + 1/(x-1)), x for all x other than 1.

Thus, you can extend the sequence of pairs with (using integers only):
(7 1/5, 6) (8 1/6, 7) (9 1/7, 8) and so on.

Your task is now to prove the apparent family I stated does indeed work.


Phillipe 92

Infinitely many :
for any integers p, q
a = p/q and b = p^2/(q*(p-q))
b - a = b/a = p/(p-q)

for instance p=16, q=2 result into a = 8, b = 64/7

64/7 - 8 = 8/7
(64/7)/8 = 8/7

> NB:- 5 1/3, should be read as 5 and 1 by 3

5 1/3 = 16/3 without any ambiguity
64/7 = 9 1/7 if you prefer this ambiguous writing, and needing a 'NB:'to explain...


Bill

"b - a = b/a" can be solved to give b = (a^2) / (a-1)
FYI: b + a = b*a = a = b / (b = 1)


Phillipe 92

Bill wrote :


> "b - a = b/a" can be solved to give b = (a^2) / (a-1)

As it seemed to be restricted to _rational_ numbers, I wrote 'a' as any rational number : p/q, and we get then the same formula...(p/q)^2/(p/q - 1 ) = p^2/(p*q - q^2) = p^2/(q*(p-q))

BTW this gives *all* rational solutions.

Restricting even more to integers results into much less solutions...
(the only solution is 4 - 2 = 4/2)

There is no interest in allowing a,b to be any _real_ number.