Yes, here is an example for that in 2 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0), such that A = (4,0). Then we have z = (4^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4) = (1, 0). The normalized vector of A does not contain a fraction.
Yes, here is an example for that in 2 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0), such that A = (4,0). Then we have z = (4^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4) = (1, 0). The normalized vector of A does not contain a fraction.
user
Here is an example for that in 3 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0,0), such that A = (4,0,0). Then we have z = (4^2 + 0^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4, 0/4) = (1, 0, 0). The normalized vector of A does not contain a fraction.

Here is an example for that in 4 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0,0,0), such that A = (4,0,0,0). Then we have z = (4^2 + 0^2 + 0^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4, 0/4, 0/4) = (1, 0, 0, 0). The normalized vector of A does not contain a fraction.

Here is an example for that in 5 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0,0,0,0), such that A = (4,0,0,0,0). Then we have z = (4^2 + 0^2 + 0^2 + 0^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4, 0/4, 0/4, 0/4) = (1, 0, 0, 0, 0). The normalized vector of A does not contain a fraction.

Here is an example for that in 6 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0,0,0,0,0), such that A = (4,0,0,0,0,0). Then we have z = (4^2 + 0^2 + 0^2 + 0^2 + 0^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4, 0/4, 0/4, 0/4, 0/4) = (1, 0, 0, 0, 0, 0). The normalized vector of A does not contain
assistant
Yes, here is an example for that in 2 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0), such that A = (4,0). Then we have z = (4^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4) = (1, 0). The normalized vector of A does not contain a fraction.
user
Now, let vector B be a vector with its initial point at the origin and terminal point at (0,4), such that B = (0,4). Then we have z = (0^2 + 4^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (0/4, 4/4) = (0, 1). The normalized vector of B does not contain a fraction.

Now, let vector C be a vector with its initial point at the origin and terminal point at (4,4), such that C = (4,4). Then we have z = (4^2 + 4^2)^(1/2) = (8^2)^(1/2) = 8^1 = 8 and with that u = (4/8, 4/8) = (1/2, 1/2). The normalized vector of C contains a fraction.

Now, let vector D be a vector with its initial point at the origin and terminal point at (4,0), such that D = (4,0). Then we have z = (4^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4) = (1, 0). The normalized vector of D does not contain a fraction.

Now, let vector E be a vector with its initial point at the origin and terminal point at (0,4), such that E = (0,4). Then we have z = (0^2 + 4^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (0/4, 4/4) = (0, 1). The normalized vector of E does not contain a fraction.

Now, let vector F be a vector with its initial point at the origin and terminal point at (4,4), such that F = (4,4). Then we have z = (4^2 + 4^2)^(1/2) = (8^2)^(1/2) = 8^1 = 8 and with that u = (4/8, 4/8) = (1/2, 1/2). The normalized vector of F contains a fraction.

Now, let vector G be a vector with its initial point at the origin and terminal point at (4,0), such that G = (4,0). Then we have z = (4^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4) = (1, 0). The normalized vector of
assistant
Yes, here is an example for that in 2 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0), such that A = (4,0). Then we have z = (4^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4) = (1, 0). The normalized vector of A does not contain a fraction.
user
Here is an example for that in 3 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0,0), such that A = (4,0,0). Then we have z = (4^2 + 0^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4, 0/4) = (1, 0, 0). The normalized vector of A does not contain a fraction.

Here is an example for that in 4 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0,0,0), such that A = (4,0,0,0). Then we have z = (4^2 + 0^2 + 0^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4, 0/4, 0/4) = (1, 0, 0, 0). The normalized vector of A does not contain a fraction.

Here is an example for that in 5 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0,0,0,0), such that A = (4,0,0,0,0). Then we have z = (4^2 + 0^2 + 0^2 + 0^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4, 0/4, 0/4, 0/4) = (1, 0, 0, 0, 0). The normalized vector of A does not contain a fraction.

Here is an example for that in 6 dimensional space:
Let vector A be a vector with its initial point at the origin and terminal point at (4,0,0,0,0,0), such that A = (4,0,0,0,0,0). Then we have z = (4^2 + 0^2 + 0^2 + 0^2 + 0^2 + 0^2)^(1/2) = (4^2)^(1/2) = 4^1 = 4 and with that u = (4/4, 0/4, 0/4, 0/4, 0/4, 0/4) = (1, 0, 0, 0, 0, 0). The normalized vector of A does not contain
assistant