39yup, u seem to be right
@prophet sir: does this proof oversee the fallacy that we were discussing that day?
Given 2 positive integers a,b satisfying a< b . Prove that among every b consecutive positive integers there are two numbers whose product is divisible by ab .
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3 Answers
11In 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.
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39
341This looks fine.
It can be shortened considerably as the number that is divisible by a and b will be divisible by its LCM = mxy. And as soumik observed any b consecutive numbers will have atleast two multiples of m. Multiplying them gives a multiple of ab.