Let $\epsilon$ be a small positive number such that $1+\epsilon$ and $3+2\epsilon$ are machine numbers but $3+2\epsilon + \epsilon^{2}$ is computed to be $3 + 2\epsilon $. Now, let the (classical) Gram-Schmidt process be applied to the three vectors $v_{1} = fl(1 + \epsilon, 1, 1)^{T} $, $v_{2} = fl(1, 1+\epsilon, 1)^{T}$, $v_{3} = fl(1, 1, 1 + \epsilon)^{T}$. Verify that the supposedly orthonormal base produced by the computer will be
$\mathbf{x_{1}=fl[1/\sqrt{3+2\epsilon}]v_{1}}$,
$\mathbf{x_{2}=fl[1/\sqrt{2}](-1,1,0)^{T}}$,
$\mathbf{x_{3}=fl[1/\sqrt{2}](-1,0,1)^{T}}$.
I am so incredibly lost because I have no idea how to perform Gram-Schmidt on numbers that look like this!
Could someone at least show me how to get $x_{2}$ ($x_{1}$ is easy enough)? If I could see how the operations with $\epsilon$'s and floats are supposed to work in calculating $x_{1}$, I could probably figure out the rest of it myself. (I assume we're supposed to solve this problem by actually performing Gram-Schmidt on $v_{1}$, $v_{2}$, and $v_{3}$; correct me if I'm wrong).
Here the calculation for $x_2:$ $$u_1=v_1, \quad <u_1,u_1>=(1+\epsilon)^2 + 1 +1 = 3+2\epsilon$$ $$u_2 = v_2 - \frac{<u_1,v_2>}{<u_1,u_1>}u_1 =(1,1+\epsilon,1)^T-\frac{(1+\epsilon)\times 1+ 1\times (1+\epsilon)+1\times 1}{3+2\epsilon}u_1= (1,1+\epsilon,1)^T-\frac{3+2\epsilon}{3+2\epsilon}u_1= (1,1+\epsilon,1)^T-(1+\epsilon,1,1)^T = (-\epsilon, \epsilon, 0)^T $$ Now $|u_2|=\sqrt{\epsilon^2 + \epsilon^2 + 0} = \epsilon \sqrt{2}$ and therefore the floating point approximation: $$x_2=\frac{u_2}{|u_2|}=fl(1/\sqrt{2}) (-1,1,0)^T$$