Total length of a normalized vector should be 1. Therefore, if it is shared to more than one dimension, the components would be less then one and therefore fractional. 

The degenerate case for one dimension is:
A = (1), then 
z = 1, u = (1). 

The case for a n-dimensional vector with n = 2 is: 
A = (3, 4),
z = sqrt( 3^2 + 4^2 ) = 5
u = (3/5, 4/5), equivalently
u = (0.6, 0.8). 

The only other case when the normalized vector does not contain fractions is when it is zero. For instance, for a four-dimensional space with n = 4:
A = (0, 0, 0, 0),
z = 0, u = (0, 0, 0, 0).
Total length of a normalized vector should be 1. Therefore, if it is shared to more than one dimension, the components would be less then one and therefore fractional. 

The degenerate case for one dimension is:
A = (1), then 
z = 1, u = (1). 

The case for a n-dimensional vector with n = 2 is: 
A = (3, 4),
z = sqrt( 3^2 + 4^2 ) = 5
u = (3/5, 4/5), equivalently
u = (0.6, 0.8). 

The only other case when the normalized vector does not contain fractions is when it is zero. For instance, for a four-dimensional space with n = 4:
A = (0, 0, 0, 0),
z = 0, u = (0, 0, 0, 0).
user
Total length of a normalized vector should be 1. Therefore, if it is shared to more than one dimension, the components would be less then one and therefore fractional. 

The degenerate case for one dimension is:
A = (1), then 
z = 1, u = (1). 

The case for a n-dimensional vector with n = 2 is: 
A = (3, 4),
z = sqrt( 3^2 + 4^2 ) = 5
u = (3/5, 4/5), equivalently
u = (0.6, 0.8). 

The only other case when the normalized vector does not contain fractions is when it is zero. For instance, for a four-dimensional space with n = 4:
A = (0, 0, 0, 0),
z = 0, u = (0, 0, 0, 0). 

Therefore, the only case when the normalized vector does not contain fractions is when it is zero.

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Q: When is a normalized vector not a fractional vector?
assistant
Total length of a normalized vector should be 1. Therefore, if it is shared to more than one dimension, the components would be less then one and therefore fractional. 

The degenerate case for one dimension is:
A = (1), then 
z = 1, u = (1). 

The case for a n-dimensional vector with n = 2 is: 
A = (3, 4),
z = sqrt( 3^2 + 4^2 ) = 5
u = (3/5, 4/5), equivalently
u = (0.6, 0.8). 

The only other case when the normalized vector does not contain fractions is when it is zero. For instance, for a four-dimensional space with n = 4:
A = (0, 0, 0, 0),
z = 0, u = (0, 0, 0, 0).
user
Total length of a normalized vector should be 1. Therefore, if it is shared to more than one dimension, the components would be less then one and therefore fractional. 

The degenerate case for one dimension is:
A = (1), then 
z = 1, u = (1). 

The case for a n-dimensional vector with n = 2 is: 
A = (3, 4),
z = sqrt( 3^2 + 4^2 ) = 5
u = (3/5, 4/5), equivalently
u = (0.6, 0.8). 

The only other case when the normalized vector does not contain fractions is when it is zero. For instance, for a four-dimensional space with n = 4:
A = (0, 0, 0, 0),
z = 0, u = (0, 0, 0, 0). 

Therefore, the only case when the normalized vector does not contain fractions is when it is zero.
assistant
Total length of a normalized vector should be 1. Therefore, if it is shared to more than one dimension, the components would be less then one and therefore fractional. 

The degenerate case for one dimension is:
A = (1), then 
z = 1, u = (1). 

The case for a n-dimensional vector with n = 2 is: 
A = (3, 4),
z = sqrt( 3^2 + 4^2 ) = 5
u = (3/5, 4/5), equivalently
u = (0.6, 0.8). 

The only other case when the normalized vector does not contain fractions is when it is zero. For instance, for a four-dimensional space with n = 4:
A = (0, 0, 0, 0),
z = 0, u = (0, 0, 0, 0).
user
Total length of a normalized vector should be 1. Therefore, if it is shared to more than one dimension, the components would be less then one and therefore fractional. 

The degenerate case for one dimension is:
A = (1), then 
z = 1, u = (1). 

The case for a n-dimensional vector with n = 2 is: 
A = (3, 4),
z = sqrt( 3^2 + 4^2 ) = 5
u = (3/5, 4/5), equivalently
u = (0.6, 0.8). 

The only other case when the normalized vector does not contain fractions is when it is zero. For instance, for a four-dimensional space with n = 4:
A = (0, 0, 0, 0),
z = 0, u = (0, 0, 0, 0). 

Therefore, the only case when the normalized vector does not contain fractions is when it is zero.

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Q: When is a normalized vector not a fractional vector?
assistant