Processing math: 3%

Tuesday, September 8, 2015

Prove that: n+1/(n+(n+2))=n, for all n\in N.

Prove that: \left\lfloor{\sqrt{n}+\dfrac{1}{\sqrt{n}+\sqrt{n+2}}}\right\rfloor= \left\lfloor{\sqrt{n}}\right\rfloor, for all n\in N.

My solution:

Step 1:

First, note that the expression inside the floor function on the left can be rewritten such that we have:

\sqrt{n}+\dfrac{1}{\sqrt{n}+\sqrt{n+2}}

=\sqrt{n}+\dfrac{1}{\sqrt{n}+\sqrt{n+2}}\cdot \dfrac{\sqrt{n}-\sqrt{n+2}}{\sqrt{n}-\sqrt{n+2}}

=\sqrt{n}+\dfrac{\sqrt{n}-\sqrt{n+2}}{n-(n+2)}

=\sqrt{n}-\dfrac{\sqrt{n}-\sqrt{n+2}}{2}

=\sqrt{n}+\dfrac{\sqrt{n+2}-\sqrt{n}}{2}

=\dfrac{\sqrt{n}+\sqrt{n+2}}{2}

Step 2:

Next, note that for all n\in N, 4\sqrt{n}\gt -2 is always true. Algebraically manipulating it such that we get:

n+4\sqrt{n}+4\gt -2+n+4

(\sqrt{n}+2)^2\gt n+2

\sqrt{n}+2\gt \sqrt{n+2}

Therefore we have:

\sqrt{n}+\sqrt{n}+2\gt \sqrt{n}+\sqrt{n+2}, which is

2\sqrt{n}+2\gt \sqrt{n}+\sqrt{n+2}

Step 3:

Note that we can set:

\sqrt{n}+\sqrt{n}\lt \sqrt{n}+\sqrt{n+2}\lt 2\sqrt{n}+2

which is

2\sqrt{n}\lt \sqrt{n}+\sqrt{n+2}\lt 2(\sqrt{n}+1)

\sqrt{n}\lt \dfrac{\sqrt{n}+\sqrt{n+2}}{2}\lt \sqrt{n}+1

Thus, we can conclude at this juncture that \left\lfloor{\dfrac{\sqrt{n}+\sqrt{n+2}}{2}}\right\rfloor=\left\lfloor{\sqrt{n}+\dfrac{1}{\sqrt{n}+\sqrt{n+2}}}\right\rfloor=\left\lfloor{\sqrt{n}}\right\rfloor and we're hence done.

1 comment:

  1. i dont Think that you can draw that conclusion simple because sqrt(x) is not necessarily an integer. för example if x € (2.3 , 3.3) then the integer part of x could either 2 or 3.
    Sébastien

    ReplyDelete