divisible

In every set of b consecutive integers, there exists a number divisible by b.

In every set of b consecutive integers, there exists atleast 1 number divisible by a.

So there exists 2 nos. whose prod is divisible by ab.

But if the numbers are equal? I mean suppose in some set of b consecutive integers, the same number happens to be divisible by both a and b. If (a,b)=1, then its done.

But let (a,b)=m. So b=mx and a=my.

Let the number that is divisible by both a and b be t=mxp=myq.

Since (x,y)=1, which gives x|q, that is, q=rx.

So t=myrx.

Since a<b, y<x which means x≥2.

Thus in any set of b consecutive integers, there exists atleast 2 naturals, both of which are divisible by m.

If 1 of them be t, let the other be t₀=md.

Thus t.t₀=m²yxrd=(my)(mx)dr=abdr

Thus proved that in any set of b consecutive integers, there exists 2, whose prod is divisible by ab.

[1]

yup, u seem to be right

@prophet sir: does this proof oversee the fallacy that we were discussing that day?