Representing the statement using Quantifiers

Question

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.

Answer 1

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.

Answer 2

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$.

Answer 3

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.