Representing the statement using Quantifiers I want to represent the statement "Some numbers are not real " using quantifiers. I have been told by my teacher that the correct way to represent this is
num(x)  : x is a number
real(x) : x is real

∃x (num(x) ^ ¬real(x)) 

This made sense, i can translate this statement into "There exist some x such that x is a number and x is not real.
But the answer i came up by myself is this one

∃x (num(x)=> ¬real(x))

In translation , There exist some x such that if x is a number then x is not real. 
I just can't get around why my answer is wrong, for some x ; if x is a number then x is not real. Doesn't that sound very similar to the statement "Some numbers are not real".
In one of the video lectures i saw  this  example which made me even more confused. 
"No dog is intelligent" 
dog(x) : x is a dog
intel(x) : x is intelligent
The representation was 

∀x (dog(x) ==> ¬intel(x))

if this representation is true, how is my representation of "Some numbers are not real" wrong. 
PS : I am just a beginner at Discrete math finding my way, please pardon me if the question doesn't meet the quality standards of the community. 
 A: Your version is wrong because "A and B" is not the same as "if A, then B".
For instance, there exists a horse H such that if H is 50 feet tall, then I win the lottery.
It is sadly not true that there exists a horse H such that H is 50 feet tall and I win the lottery.
More pointedly, (A implies B) is true when A is false. Therefore, "there exists X such that if X is a number, then X is not real" is true when X, for instance, is a banana.
A: One strategy is to write up the contra-positive statement. "All numbers are real", and then negate it. Something like
$$ \neg( \forall x (num(x)\rightarrow real(x)) $$
Remember that $ p \rightarrow q  \iff  \neg p \vee q$.
A: Restricted existentials need conjunctions. 'Some $A$s are not $B$s" says that something which is an $A$ is also not a $B$, so $\exists x(Ax \land \neg Bx)$.
$\exists x(Ax \to \neg Bx)$ is true if there is something which satisfies the condition $Ax \to \neg Bx$, and anything that doesn't satisfy $A$ will make the antecedent of the condition false  and hence make the condition true. So $\exists x(Ax \to \neg Bx)$ is true if something isn't $A$, which isn't what you want at all.
Any elementary logic text will explain how to express restricted quantifications and stop you making this very elementary mistake in translation -- e.g. look for Paul Teller's excellent Modern Logic Primer, now freely available online, whose chapters on translation with quantifiers are particularly good.
