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1.

Define $$\{x\} = x-\lfloor x \rfloor$$. That is to say, {x} is the "fractional part" of x. For example, if you were to expand a positive number x as a decimal, {x} is the stuff after the decimal point. For example $$\left\{\frac{3}{2}\right\} = 0.5$$ and $$\{\pi\} = 0.14159\dots$$

Now, using the above definition, determine if the function below is increasing, decreasing, even, odd, and/or invertible on its natural domain:

$$f(x) = \lfloor x \rfloor - \left\{ x \right\}$$

For each property, write inCreasing, Decreasing, Even, Odd, inVertible in that order (alphabetical). For example, if the function is increasing, odd, and invertible, submit "COV". If the function is none of the above, submit "NONE".

2.

Write $$\frac{\sqrt{2}}{\sqrt{2} + \sqrt{3} - \sqrt{5}} - \frac{1}{2}$$ with a rational denominator.

Apr 22, 2021

#1
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One question per post, thanks

Apr 22, 2021
#2
+8461
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$$\quad\dfrac{\sqrt 2}{\sqrt 2 + \sqrt 3 - \sqrt 5} - \dfrac12\\ = \dfrac{\sqrt2 \left(\sqrt2 + \sqrt 3 + \sqrt 5\right)}{(\sqrt 2 + \sqrt 3)^2 - \sqrt5^2} - \dfrac12\\ =\dfrac{\sqrt 2 \left(\sqrt 2 + \sqrt 3 + \sqrt 5\right)}{2\sqrt 6} - \dfrac12\\ = \dfrac{\sqrt 2 + \sqrt 3 + \sqrt 5}{2 \sqrt 3} - \dfrac{\sqrt 3}{2\sqrt 3}\\ =\dfrac{\sqrt 2 + \sqrt 5}{2 \sqrt 3}\\ = \dfrac{\sqrt 6 + \sqrt{15}}{6}$$

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Apr 22, 2021