Question

Determine the truth value of each of these statements if thedomainofeachvariableconsistsofallrealnumbers.

a) ∀ x ∃ y(x 2 = y)
b) ∀ x ∃ y(x = y 2 )
c) ∃ x ∀ y(xy = 0 )
d) ∃ x ∃ y(x + y = y + x)
e) ∀ x(x = 0 →∃ y(xy = 1 ))
f) ∃ x ∀ y(y = 0 → xy = 1 )
g) ∀ x ∃ y(x + y = 1 )
h) ∃ x ∃ y(x + 2 y = 2 ∧ 2 x + 4 y = 5 )
i) ∀x∃y(x + y = 2 ∧ 2x − y = 1)
j) ∀x∀y∃z(z = (x + y)/2)

Answer 1

Answer:

a)TRUE

b)FALSE

c)TRUE

d)FALSE

e)TRUE

f)TRUE

g)TRUE

h)FALSE

i)FALSE

j)TRUE

Step-by-step explanation:

a) For every x there is y such that  $x^2=y$:

 TRUE

This statement is true, because for every real number there is a square         number of that number, and that square number is also a real number. For example, if we take 6.5, there is a square of that number and it equals 39.0625.

b) For every x there is y such that  $x=y^2$:

 FALSE

For example, if x = -1, there is no such real number so that its square equals -1.

c) There is x for every y such that xy = 0

 TRUE

If we put x = 0, then for every y it will be xy=0*y=0

d)There are x and y such that $x+y\neq y+x$

 FALSE

There are no such numbers. If we rewrite the equation we obtain an incorrect statement:

                                   $x+y \neq y+x\\x+y - y-y\neq 0\\0\neq 0$

e)For every x, if   $x \neq 0$  there is y such that xy=1:

 TRUE

The statement is true. If we have a number x, then multiplying x with 1/x (Since x is not equal to 0 we can do this for ever real number) gives 1 as a result.

f)There is x for every y such that if $y\neq 0$ then xy=1.

TRUE

The statement is equivalent to the statement in e)

g)For every x there is y such that x+y = 1

TRUE

The statement says that for every real number x there is a real number y such that x+y = 1, i.e. y = 1-x

So, the statement says that for every real umber there is a real number that is equal to 1-that number

h) There are x and y such that

                                  $x+2y = 2\\2x+4y = 5$

We have to solve this system of equations.

From the first equation it yields x=2-2y and inserting that into the second equation we have:

                                   $2(2-2y)+4y=5\\4-4y+4y=5\\4=5$

Which is obviously false statement, so there are no such x and y that satisfy the equations.

FALSE

i)For every x there is y such that

                                     $x+y=2\\2x-y=1$

We have to solve this system of equations.

From the first equation it yields $x=2-y$  and inserting that into the second equation we obtain:

                                        $2(2-y)-y=1\\4-2y-y=1\\4-3y=1\\-3y=1-4\\-3y=-3\\y=1$

Inserting that back to the first equation we obtain

                                            $x=2-1\\x=1$

So, there is an unique solution to this equations:

x=1 and y=1

The statement is FALSE, because only for x=1 (and not for every x) exists y (y=1) such that

                                         $x+y=2\\2x-y=1$

j)For every x and y there is a z such that

                                      $z=\frac{x+y}{2}$

TRUE

The statament is true for all real numbers, we can always find such z. z is a number that is halway from x and from y.