In the questions below P(x,y) means “x and y are real numbers such that x + 2y = 5”.
a) Determine whether the statement ∃x∀yP(x,y) is true.
b) Negate the statement ∃x∀yP(x,y).
a) Choose any x, you can easily find a y that makes P(x,y) false.
b) ∃ and ∀ switch with negation;
∀x∃y ~P(x,y)
a) Well, let me think. If we have P(x,y) which means x + 2y = 5, then let's see if there is an x that works for all y. If we plug in x = 1, we get 1 + 2y = 5, which simplifies to 2y = 4, and y = 2. So for x = 1, we have a valid solution. Therefore, the statement ∃x∀yP(x,y) is true.
b) Alright, let's negate the statement ∃x∀yP(x,y). In logical terms, it becomes ¬(∃x∀yP(x,y)). Now, let me put on my thinking cap again. Negating this statement means there doesn't exist an x such that for all y, P(x,y) holds true. In other words, there is no x+y combo that satisfies x + 2y = 5. So, the negation of ∃x∀yP(x,y) would be ∀x∃y(¬P(x,y)). It's like saying "For all x, there is a y such that x + 2y is not equal to 5."
To answer the questions, let's break down the statements step-by-step:
a) Determine whether the statement ∃x∀yP(x,y) is true.
To determine the truth of this statement, we need to check if there exists a value for x such that the equation x + 2y = 5 holds true for all real numbers y.
To find a possible value for x, we can rearrange the equation:
x = 5 - 2y
Now, we can substitute any value for y into the equation and find an x that satisfies it. Let's try a few values for y:
When y = 0:
x = 5 - 2(0) = 5
So, for y = 0, x = 5, satisfying the equation.
When y = 1:
x = 5 - 2(1) = 3
So, for y = 1, x = 3, satisfying the equation.
When y = 2:
x = 5 - 2(2) = 1
So, for y = 2, x = 1, satisfying the equation.
From these examples, we can see that for any real number y, there exists a corresponding x that satisfies the equation x + 2y = 5. Therefore, the statement ∃x∀yP(x,y) is true.
b) Negate the statement ∃x∀yP(x,y).
To negate the statement, we need to change the quantifiers (∃ and ∀) and negate the expression P(x,y).
The negation of ∃x is ∀x, and the negation of ∀y is ∃y. The negation of P(x, y) is "not P(x, y)" or "¬P(x, y)".
Therefore, the negation of the statement ∃x∀yP(x,y) is ∀x∃y¬P(x,y).
To determine the truth value of the statement ∃x∀yP(x,y), we need to understand the meaning of the symbols and the logical structure of the statement.
In this case, ∃ represents the existential quantifier, which means "there exists." ∀ represents the universal quantifier, which means "for all."
The statement ∃x∀yP(x,y) can be read as "There exists a real number x such that for all real numbers y, x + 2y = 5." In other words, we are trying to find a single value of x that satisfies the equation x + 2y = 5 for all values of y.
To determine if the statement is true, we need to find such an x that satisfies the equation for all possible y.
a) Determining the truth value of ∃x∀yP(x,y):
To find such an x, we can solve the equation x + 2y = 5 for x in terms of y. Rearranging the equation, we have:
x = 5 - 2y
Now, substitute x = 5 - 2y into the original equation:
(5 - 2y) + 2y = 5
Simplifying the equation, we have:
5 - 2y + 2y = 5
5 = 5
The equation holds true regardless of the value of y. This means that y can take any real value, and the equation will still be satisfied.
Since we have found that there is a single value of x (5 - 2y) that satisfies the equation for all possible values of y, we can conclude that the statement ∃x∀yP(x,y) is true.
b) Negating the statement ∃x∀yP(x,y):
To negate the statement ∃x∀yP(x,y), we need to negate each part separately and change the quantifiers accordingly.
The negation of ∃x is ∀x, and the negation of ∀y is ∃y. The negation of P(x,y) is ¬P(x,y), which means "not P(x,y)" or "x + 2y ≠ 5".
Therefore, the negation of ∃x∀yP(x,y) is ∀x∃y(¬P(x,y)), which can be read as "For all real numbers x, there exists a real number y such that x + 2y ≠ 5."