Note: Vectors like \([x_{1}, x_{2}]\) are supposed to be displayed vertically, not horizontally. I just haven’t figured out how yet.
Show that the set
\[W = \{[x_{1}, x_{2}] | 3x_{1}-5x_{2} = 12\}\] fails the Additive Closure and Scalar Closure property
Suppose \([y_{1}, y_{2}]\) is a vector in \(W\) as well as \([x_{1},x_{2}]\)
\(x\) must satisfy \(3x_{1}-5x_{2}=12\) and \(y\) must satisfy \(3y_{1}-5y_{2}=12\)
To satisfy the AC Property, \(x + y\) is in \(W\)
\([x_{1}, x_{2}]\) + \([y_{1},y_{2}]\) = \([x_{1} + y_{1}, x_{2} + y_{2}]\)
\(3(x_{1} + y_{1}) - 5(x_{2} + y_{2}) = 12\)
\(3x_{1} + 3y_{1} - 5x_{2} - 5y_{2} = 12\)
\((3x_{1} - 5x_{2}) + (3y_{1} - 5y_{2}) = 12\)
\(12 + 12 \neq 12\) So AC propery fails
To satisfy the SC Property, \(\alpha[x_{1}, x_{2}]\) = \([\alpha x_{1}, \alpha x_{2}]\)
so
\(\alpha 3x_{1} - \alpha 5x_{2}\) = 12
\(\alpha (3x_{1} - 5x_{2})\) = 12
\(\alpha 12 \neq 12\) unless \(\alpha\) = 1? I’m assuming since all numbers except 1 don’t work, the property isn’t satified.