There are given the ratios a/b = b/c = c/a . Demonstrate that a=b=c !
a/b = b/c = c/a
Let a/b = x ==> a = bx ...(1)
==> b/c = x==> b = cx .....(2)
==> c/a = x ==> c = ax .....(3)
Multiply (1) (2) and (3)
==> abc = abc x
==> x = 1
==> a/b = x = 1
==> a = b = cx = c
==> a= b = c
a/b=b/c=c/a = k say,
Then a = bk, b = ck and c = ak
Or a = bk = (ck)k = (ak)kk. So a= ak^3. So k = 1.
Therefore a=bk =b = ck = c
Let’s assume that the value of the three quotients is the constant k, so that:
a/b=k, b*k=a (1)
b/c=k, c*k=b (2)
c/a=k, a*k=c (3)
If we are multiplying (1),(2),(3) then the result will be:
b*k* c*k* a*k=a*b*c
a*b*c*k^3= a*b*c (4)
Because a,b,c different from 0, then their product, a*b*c, is also not cancelling.
Because of this, we’ll divide the relation (4), by the product a*b*c and we'll get:
k^3=1, that means that k=1, so a/b=b/c=c/a=k=1
We’ll substitute k=1 and the relation (1) will become: b*1=a , so b=a
We’ll substitute k=1 and the relation (2) will become: c*1=b , so b=c, but b=a so, from transitivity relation results that c=a.
In the end we'll get: