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
:
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
:
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](https://tex.z-dn.net/?f=x%2By%5Cneq%20y%2Bx)
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](https://tex.z-dn.net/?f=x%2By%20%5Cneq%20y%2Bx%5C%5Cx%2By%20-%20y-y%5Cneq%200%5C%5C0%5Cneq%200)
e)For every x, if
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
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](https://tex.z-dn.net/?f=x%2B2y%20%3D%202%5C%5C2x%2B4y%20%3D%205)
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](https://tex.z-dn.net/?f=2%282-2y%29%2B4y%3D5%5C%5C4-4y%2B4y%3D5%5C%5C4%3D5)
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](https://tex.z-dn.net/?f=x%2By%3D2%5C%5C2x-y%3D1)
We have to solve this system of equations.
From the first equation it yields
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](https://tex.z-dn.net/?f=2%282-y%29-y%3D1%5C%5C4-2y-y%3D1%5C%5C4-3y%3D1%5C%5C-3y%3D1-4%5C%5C-3y%3D-3%5C%5Cy%3D1)
Inserting that back to the first equation we obtain
![x=2-1\\x=1](https://tex.z-dn.net/?f=x%3D2-1%5C%5Cx%3D1)
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](https://tex.z-dn.net/?f=x%2By%3D2%5C%5C2x-y%3D1)
j)For every x and y there is a z such that
![z=\frac{x+y}{2}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx%2By%7D%7B2%7D)
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.