Notes: `(x/y)^(n + 1) = 1 and 0

Tuesday, May 13, 2014

$\displaystyle{{\left(\frac{{x}}{{y}}\right)}}^{{{n}+{1}}}={1}{\quad\text{and}\quad}{0}$

You need to use mathematical induction to prove the inequality, hence, you need to perform the following two steps, such that:

Step 1: Basis: Prove that the statement holds for n = 1

If x<y then $\displaystyle{{\left(\frac{{x}}{{y}}\right)}}^{{2}}\le{{\left(\frac{{x}}{{y}}\right)}}^{{1}}$ holds

Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds.

$\displaystyle{P}{\left({k}\right)}:{{\left(\frac{{x}}{{y}}\right)}}^{{{k}+{1}}}\le{{\left(\frac{{x}}{{y}}\right)}}^{{k}}$ holds

$\displaystyle{P}{\left({k}+{1}\right)}:{{\left(\frac{{x}}{{y}}\right)}}^{{{k}+{2}}}\le{{\left(\frac{{x}}{{y}}\right)}}^{{{k}+{1}}}$

$\displaystyle{{\left(\frac{{x}}{{y}}\right)}}^{{{k}+{1}}}\cdot{\left(\frac{{x}}{{y}}\right)}\le{{\left(\frac{{x}}{{y}}\right)}}^{{{k}+{1}}}\Rightarrow\frac{{x}}{{y}}\le{1}\Rightarrow{x}\le{y}$ true

Hence, since both the basis and the inductive step hold, the statement $\displaystyle{P}{\left({n}\right)}:{{\left(\frac{{x}}{{y}}\right)}}^{{{n}+{1}}}\le{{\left(\frac{{x}}{{y}}\right)}}^{{n}}$ holds for all indicated values of n.

Notes

Tuesday, May 13, 2014

$\displaystyle{{\left(\frac{{x}}{{y}}\right)}}^{{{n}+{1}}}={1}{\quad\text{and}\quad}{0}$

No comments:

Post a Comment

How does the choice of details set the tone of the sermon?