Piecewise function

Ummmmm, I guess I will just simplify???

\(z=\frac{{\sqrt{3}}^2-2{\sqrt{2}}^2}{\sqrt{3}-2\sqrt{2}}\)

Then simplify squared terms

\(z=\frac{3-2*2}{\sqrt{3}-2\sqrt{2}}\)

Simplify further

\(z=\frac{-1}{\sqrt{3}-2\sqrt{2}}\)

Now, lets see if \(\sqrt{3}-2\sqrt{2}\), the denominator, is larger than 1.

round sqrt(3) and sqrt(2) into, 1.7 and 1.4, respectively.

1.7 - 2(1.4) = -1.1

So we have \(z=\frac{-1}{-1.1}\)

\(\frac{1}{1.1}\)

The floor function of that is 0, so that should be the answer???

Piecewise function

If  \(z = \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} } \)  find \(\lfloor z \rfloor\).

The fractional part denoted by  \(\{x\}\) for real x and defined by the formula \( \{x\}=x-\lfloor x\rfloor\).

Source: https://en.wikipedia.org/wiki/Floor_and_ceiling_functions

\(\begin{array}{|rcll|} \hline z &=& \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} } \quad | \quad \{ \sqrt{3} \} = \sqrt{3}- \lfloor \sqrt{3}\rfloor,\ \{ \sqrt{2} \} = \sqrt{2}- \lfloor \sqrt{2}\rfloor \\\\ &=& \dfrac{ \left(\sqrt{3}- \lfloor \sqrt{3}\rfloor\right)^2 - 2 \left(\sqrt{2}- \lfloor \sqrt{2}\rfloor\right)^2 }{ \left(\sqrt{3}- \lfloor \sqrt{3}\rfloor\right) - 2 \left(\sqrt{2}- \lfloor \sqrt{2}\rfloor\right) } \quad | \quad \lfloor \sqrt{3}\rfloor = \lfloor \sqrt{2}\rfloor = 1 \\\\ &=& \dfrac{ \left(\sqrt{3}- 1\right)^2 - 2 \left(\sqrt{2}- 1\right)^2 }{ \left(\sqrt{3}- 1\right) - 2 \left(\sqrt{2}- 1\right) } \\\\ &=& \dfrac{ 3-2\sqrt{3} + 1 -2(2-2\sqrt{2}+1) } { \sqrt{3}-1-2\sqrt{2} +2 } \\\\ &=& \dfrac{ 3-2\sqrt{3} + 1 -4 +4\sqrt{2}-2 } { \sqrt{3}-2\sqrt{2} +1 } \\\\ &=& \dfrac{ -2\sqrt{3} +4\sqrt{2}-2 } { \sqrt{3}-2\sqrt{2} +1 } \\\\ &=& \dfrac{ -2\left( \sqrt{3} -2\sqrt{2}+1\right) } { \left(\sqrt{3}-2\sqrt{2} +1\right) } \\\\ &=& -2 \\ \hline \mathbf{ \lfloor z\rfloor } &=& \mathbf{ -2 } \\ \hline \end{array}\)