dadagreat wrote:

m and n are two consecutive positive integers. Is m > n ?

1. m-1 and n+1 are consecutive positive integers

2. m is an even integer

I am not able to understand S1. if m and n are consecutive +ve integers, how can m-1 and n+1 be consecutive +ve integers?

The explanation just says "S1 holds only when m=n+1 . It is sufficient." - couldnt make sense out of it.

can someone help?

Given : m and n are consecutive integers

Now there are two possibilities only. Obviously you cant make out because

m>n or n>m

Take any 2 numbers for first possibility where m=n+1. So no.s here we have are n and n+1

with S1=> m-1= n+1-1=n -----(1)

n+1= n+1---(2)

1 and 2 are consecutive from first possibility. This means if you know that m+1 and n-1 are consecutive, then you know that m and n were also consecutive, with m>n.

But say, if they are not, which will happen in case of another possibility, then this cant be deducted. If we go to another possibility and do same analysis we wont find this true. I have given that also below, however not needed.

With another possibility, m=n-1 also. we have n-1 and n.

Now S1=> m-1= n-1-1= n-2----(1)

n+1 n+1----(2)