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

a)TRUE

b)FALSE

c)TRUE

d)FALSE

e)TRUE

f)TRUE

g)TRUE

h)FALSE

i)FALSE

j)TRUE

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\\eq y+x`

FALSE

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

`x+y \\eq y+x\\x+y - y-y\\eq 0\\0\\eq 0`

e)For every x, if
`x \\eq 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\\eq 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=(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.