I have to prove that f = {(x,n) e R x Z | n =< x < n + 1} a function is.
I started with saying (x,n1) and (x,n2) are both elements of the function. This means n1 =< x < n1 + 1 (1) and n2 =< x < n2 + 1 (2). I substracted n1 from (1) and n2 from (2) this gives 0 =< x - n1 < 1 (3) and 0 =< x - n2 < 1 (4) when I substract (3) from (4), this gives 0 =< n1 - n2 < 0 (5). This is impossible since a number can't be both greater than or equal to 0 and smaller than 0. I wanted to work towards n1 - n2 = 0. Can someone help me find the mistake I made?
Thanks!
We cannot subtract inequalities!
Consider for example $1<2,\ 0<2$, if you subtract them, you would receive $1<0$ which is false.
It's not working because addition of inequalities does work, and subtraction is $a-b=a+(-b)$, while multiplication by $-1$ reverses the inequality to be subtracted.
If we carefully add appropriate forms of the inequalities instead: $$0\le (x-n_1)<1\\ -1<(n_2-x)\le 0$$ then we do get $-1<(n_2-n_1)<1$, implying $n_2-n_1=0$ as these are integers.