Question

Determine the truth value of each of these statements if the domain for all variables consists of all real numbers.

a) \(\forall x\exists y\left( {{x^2} = y} \right)\) b) \(\forall x\exists y\left( {x = {y^2}} \right)\)

c) \(\exists x\forall y\left( {xy = 0} \right)\) d) \(\exists x\exists y\left( {x + y \ne y + x} \right)\)

e) \(\forall x\left( {x \ne 0 \to \exists y\left( {xy = 1} \right)} \right)\) f) \(\exists x\forall y\left( {y \ne 0 \to xy = 1} \right)\)

g) \(\forall x\exists y\left( {x + y = 1} \right)\) h) \(\exists x\exists y\left( {x + 2y = 2 \wedge 2x + 4y = 5} \right)\)

i) \(\forall x\exists y\left( {x + y = 2 \wedge 2x - y = 1} \right)\) j) \(\forall x\forall y\exists z\left( {z = \left( {x + y} \right)/2} \right)\)

Short answer

The Truth values can be Determine.

Step-by-step solution

To finding the truth values for \(\forall x\exists y\left( {{x^2} = y} \right)\)

The Statement “\(\forall x\exists y\left( {{x^2} = y} \right)\)”

The domain of each variable consists of all real numbers.

The Statement “\(\forall x\exists y\left( {{x^2} = y} \right)\)”, is true for any value of x and there exist a y value

For Example:

\( \Rightarrow x = 1,y = 1 \Rightarrow \left( {{x^2} = y} \right)\left( {true} \right)\)

\( \Rightarrow x = 2,y = 4 \Rightarrow \left( {{x^2} = y} \right)\left( {true} \right)\)

As a result, The Truth value of the statement “\(\forall x\exists y\left( {{x^2} = y} \right)\)” is True.

To finding the truth values for \(\forall x\exists y\left( {x = {y^2}} \right)\)

The Statement “\(\forall x\exists y\left( {x = {y^2}} \right)\)”

The domain of each variable consists of all real numbers.

The Statement “\(\forall x\exists y\left( {x = {y^2}} \right)\)”, is False Statement

For Example:

\( \Rightarrow x = - 2,y = \sqrt { - 2} \)(not a real number)

As a result, The Truth value of the statement “\(\forall x\exists y\left( {x = {y^2}} \right)\)” is False.

To finding the truth values for \(\exists x\forall y\left( {xy = 0} \right)\)

The Statement “\(\exists x\forall y\left( {xy = 0} \right)\)”

The domain of each variable consists of all real numbers.

The Statement “\(\exists x\forall y\left( {xy = 0} \right)\)”, is true Statement

For Example:

\( \Rightarrow for\)\(x = any\)value, y=0\( \Rightarrow xy = 0\)

As a result, The Truth value of the statement “\(\exists x\forall y\left( {xy = 0} \right)\)” is True.

To finding the truth values for \(\exists x\exists y\left( {x + y \ne y + x} \right)\)

The Statement “\(\exists x\exists y\left( {x + y \ne y + x} \right)\)”

The domain of each variable consists of all real numbers.

The Statement “\(\exists x\exists y\left( {x + y \ne y + x} \right)\)”, is False Statement

For Example:

\( \Rightarrow For\)any values of\(x,y;x + y = y + x\)

As a result, The Truth value of the statement “\(\exists x\exists y\left( {x + y \ne y + x} \right)\)” is False.

To finding the truth values for \(\forall x\left( {x \ne 0 \to \exists y\left( {xy = 1} \right)} \right)\)

The Statement “\(\forall x\left( {x \ne 0 \to \exists y\left( {xy = 1} \right)} \right)\)”

The domain of each variable consists of all real numbers.

The Statement “\(\forall x\left( {x \ne 0 \to \exists y\left( {xy = 1} \right)} \right)\)”, is true Statement

For Example:

\( \Rightarrow for\)\(x = 2\)There exits y=1/2\( \Rightarrow xy = 1\)

As a result, The Truth value of the statement “\(\forall x\left( {x \ne 0 \to \exists y\left( {xy = 1} \right)} \right)\)” is True.

To finding the truth values for \(\exists x\forall y\left( {y \ne 0 \to xy = 1} \right)\)

The Statement “\(\exists x\forall y\left( {y \ne 0 \to xy = 1} \right)\)”

The domain of each variable consists of all real numbers.

The Statement “\(\exists x\forall y\left( {y \ne 0 \to xy = 1} \right)\)”, is true Statement

For Example:

\( \Rightarrow for\)\(y = 2\)There exits x=1/2\( \Rightarrow xy = 1\)

As a result, The Truth value of the statement “\(\exists x\forall y\left( {y \ne 0 \to xy = 1} \right)\)” is True.

To finding the truth values for \(\forall x\exists y\left( {x + y = 1} \right)\)

The Statement “\(\forall x\exists y\left( {x + y = 1} \right)\)”

The domain of each variable consists of all real numbers.

The Statement “\(\forall x\exists y\left( {x + y = 1} \right)\)”, is true Statement Because for any value x there exit y

For Example:

\( \Rightarrow for\)\(x = 2\)There exits y=-1\( \Rightarrow x + y = 1\)

As a result, The Truth value of the statement “\(\forall x\exists y\left( {x + y = 1} \right)\)” is True.

To finding the truth values for \(\exists x\exists y\left( {x + 2y = 2 \wedge 2x + 4y = 5} \right)\)

The Statement “\(\exists x\exists y\left( {x + 2y = 2 \wedge 2x + 4y = 5} \right)\)”

The domain of each variable consists of all real numbers.

The Statement “\(\exists x\exists y\left( {x + 2y = 2 \wedge 2x + 4y = 5} \right)\)”, is False Statement. Because non x, y values exist to make the statement true

For Example:

\(x = 2,y = 0 \Rightarrow x + 2y = 2\& 2x + 4y \ne 5\)

As a result, The Truth value of the statement “\(\exists x\exists y\left( {x + 2y = 2 \wedge 2x + 4y = 5} \right)\)” is False.

To finding the truth values for \(\forall x\exists y\left( {x + y = 2 \wedge 2x - y = 1} \right)\)

The Statement “\(\forall x\exists y\left( {x + y = 2 \wedge 2x - y = 1} \right)\)”

The domain of each variable consists of all real numbers.

The Statement “\(\forall x\exists y\left( {x + y = 2 \wedge 2x - y = 1} \right)\)”, is False Statement. Because non x, y values exist to make the statement true

For Example:

\(x = 2,y = 0 \Rightarrow x + 2y = 2\& 2x - y \ne 1\)

As a result, The Truth value of the statement “\(\forall x\exists y\left( {x + y = 2 \wedge 2x - y = 1} \right)\)” is False.

To finding the truth values for \(\forall x\forall y\exists z\left( {z = \left( {x + y} \right)/2} \right)\)

The Statement “\(\forall x\forall y\exists z\left( {z = \left( {x + y} \right)/2} \right)\)”

The domain of each variable consists of all real numbers.

The Statement “\(\forall x\forall y\exists z\left( {z = \left( {x + y} \right)/2} \right)\)”, is true Statement.

There exist z for all x, y values

For Example:

\( \Rightarrow for\)\(x = 2,y = 0 \Rightarrow z = \left( {x + y} \right)/2 = 1\left( {true} \right)\)

As a result, The Truth value of the statement “\(\forall x\forall y\exists z\left( {z = \left( {x + y} \right)/2} \right)\)” is True.