Negation of ∀ a ∈ Z, ∃ b ∈ Z such that ab ≤ 1

Question

∀ a ∈ Z, ∃ b ∈ Z such that ab ≤ 1 Which Negates to ∃ a ∈ Z, ∀ b ∈ Z such that ab ≥ 1

Aren't both the statement and negation true? Since in the list of all integers there will be a negative which can lower or raise above zero?

Answer 1

Note that for your negation, it should be $ab>1$.

A counter example for your negation is to let $b=0$ regardless of how you pick your $a$.

Answer 2

The statement is true and the negation is false.

And the negation should be: ∃ a ∈ Z, ∀ b ∈ Z such that ab > 1

The same proof for the statement(prove it is true) and for the negation(prove it is false):

If $a > 0$, then $b < 0$, then $ab< 0 < 1$.

If $a = 0$, then $ab = 0 < 1$.

If $a < 0$, then $b >0$, then $ab <0<1$.

Answer 3

Close.   The negation of $~∀ a ∈ Z, ∃ b ∈ Z : ab ≤ 1~$ is $~∃ a ∈ Z, ∀ b ∈ Z : ab > 1~$   (Note: "strictly greater than", not "at least").

Aren't both the statement and negation true? Since in the list of all integers there will be a negative which can lower or raise above zero?

No; the order of quantification is significant.

The first statement says that "for all integers, $a$, there is some integer, $b$, such that their product is at most one".   That can be a different $b$, but their is (at least) one $b$ for each $a$ that their is.   Now witness that for any $a$ you can use $-a$ as the $b$ and indeed $a(-a)\leq 1$.   So the statement is true.

The second statement says that "there is some integer, $a$ which for every integer, $b$, makes the product strictly greater than one".   That must be the same $a$ for every $b$ that there is.   Can you supply a single witness, $a$, which, when multiplied by any integer (for instance $b=0$) is strictly greater than one?   Neither can I.   The second statement is false.