Question

Which of the following are true in the universe of all real numbers?

(a) (∃x)(∀y)(x+y=0).
(b) (∃!y)((y<0)∧(y+3>0)).

Answer 1

The first statement (a) is true because for any real number y, an x exists that is the additive inverse of y, thus x+y=0. The second statement (b) is false because there are infinitely many real numbers between -3 and 0 that would satisfy the condition, not just one.

Step-by-step explanation:

The question involves understanding logical statements and inequalities in the context of real numbers. Let's analyze each statement separately:

• (a) (∃x)(∀y)(x+y=0): This statement means that there exists a number x such that for any number y, the sum of x and y equals 0. This is true because for every real number y, we can choose x to be the additive inverse of y, namely x = -y, which satisfies the equation x+y=0.
• (b) (∃!y)((y<0)∧(y+3>0)): This statement asserts that there exists exactly one number y which is less than zero and such that y+3 is greater than zero.
• This is false because if y is less than 0, adding 3 to y would make it positive only if y is greater than -3. Any real number between -3 and 0 would satisfy this condition, so there isn't just one unique y that fulfills the criteria.