I remember a high school book I read a long time ago explains that every statement in the form of $\forall x\in D,P(x)$ can be turned into an implication.
For example, isn't
"For every real number $x$, $x^2$ is non-negative."
equivalent to
"If $x$ is a real number, then $x^2$ is non-negative."?
However, I came across a wikipedia article about universal quantifier that says something about bounded quantifier, which I haven't heard before. Then I thought, and confirmed, that bounded quantifier can actually be written as unbounded quantifier, and that the statement I read in the book I said above is actually a bounded version.
For example, $\forall x\in D,P(x)$ becomes $\forall x,[x\in D \implies P(x)]$ My question is: was what I read in the book correct? That every bounded universal statement can be converted into an implication? Another one is: Can or can't every (unbounded) universal statement become implication?
I ask this because I am confused. I thought every universal statement can become an implication, the wikipedia article and the fact that I cannot find (google) any other article about my thought make me want to post this question. Besides, don't we use arguably the same starting premise when we want to prove a universal statement and an implication? Both use "Let x be ..." or "Assume that ..." or "Take arbitrary x ..." (the latter especially is used most often when facing a universal statement) as starting premise, don't they?
Thanks for the answer. ^^
ADDITION: Why can't $\forall x\in D,P(x)$ just become $x\in D \implies P(x)$ instead of $\forall x,[x\in D \implies P(x)]$?
$\forall x \in D \colon P(x)$ is simply an abbreviation for the formula $\forall x ( x \in D \implies P(x))$ -- there really isn't more to it. You will find this convention in any decent textbook that covers the basics of first order logic.
And you can't replace it by $x \in D \implies P(x)$ for the simple reason that the latter is not a sentence -- it has unbounded free variables (namely $x$). Hence it's not equivalent to the sentence $\forall x \in D \colon P(x)$.