can somebody explain this .i don't understand the way used by Matt sir

http://www.sosmath.com/CBB/viewtopic.php?p=214676&highlight=#214676

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- Mar 6th 2011, 10:45 AMayushdadhwalRange
can somebody explain this .i don't understand the way used by Matt sir

http://www.sosmath.com/CBB/viewtopic.php?p=214676&highlight=#214676

- Mar 6th 2011, 01:12 PMemakarov
Let $\displaystyle f(x)=\lfloor x^2\rfloor-\lfloor x\rfloor^2$. Then $\displaystyle \mathop{\mbox{range}}(f)\subseteq\mathbb{Z}$. To prove the converse inclusion ($\displaystyle \mathbb{Z}\subseteq\mathop{\mbox{range}}(f)$), one must show that every integer $\displaystyle n$ is in the range of $\displaystyle f$. For this, one needs to find a pre-image of every integer. Matt showed that $\displaystyle f(n+1/2)=n$.

- Mar 6th 2011, 05:25 PMayushdadhwal
sir my basic question depends on domain(INPUT).if we look at f(x) then its domain is all real numbers (INPUT) and its range is Z(OUTPUT).Then how can we straight forward choose x=n+1/2.here x means part of my domain and domain can be any real number.therefore i can take any value of x let x=1/3.now how can i choose x=n+1/2.perhaps sir you are able to understand my confusion.

- Mar 7th 2011, 02:18 AMemakarov
Consider the claim: "There exists a natural number p such that both p and p + 2 are prime". This claim is true. To prove it, consider p = 5; then 5 and 7 are prime, as required. This finished the proof. One does not have to prove that p and p + 2 are prime for

*all*natural p. One also does not have to find all possible p such that p and p + 2 are prime. One does not even have to know if there are infinitely many such p's (this is called Twin Prime Conjecture, and it is still unsolved). All is needed to prove a claim that starts with "there exists an x such that A(x)" (where A is some property) is to find a single example x that makes A(x) true.

Now, in your case, the claim is, "For every integer n, there exists a real x such that f(x) = n". A proof starts by fixing an arbitrary n. Then we need to find a single x such that f(x) = n. We don't have to find all such x's; we don't have to consider x's such that f(x) is not n. We are interested in only one thing: finding a single x such that f(x) = n. It turns out that x = n + 1/2 works because f(n + 1/2) = n.