Proof that for all real numbers a, |-a| = |a|

We can think of the absolute value of a number as its distance from 0. So, the absolute value of a, which is denoted by $latex |a|$ is always greater than 0. In this post, we are going to prove that for all real numbers a, |-a| = |a|.

There are two possible cases: $latex a \geq 0$ and $latex b < 0$. (i) For $latex a \geq 0$, $latex since - a \leq 0$ $latex |-a| = -(-a) = a$ (since a is negative, we negate it to make it positive) $latex |a| = a$ (ii) For $latex a < 0$, since $latex - a > 0$,

$latex |-a|$ = $latex -a$ (since a is negative, we negate it to make it positive)
$latex |a| = -a$

By (i) and (ii), for any real number a,

$latex |-a| = |a|$.

Derivation of the Equation of the Parabola

A circle is a locus of points  whose distance from a fixed point is a constant. A parabola can also be described as a locus of points whose distance from a fixed point and a fixed line not passing through that point is a constant. An example of a parabola is shown below.

In the figure below, point $latex F$ is called the focus of the parabola and line $latex l$ is called its directrix. The vertex of the parabola is at the origin O and the $latex x$-axis the perpendicular bisector of $latex FH$. If we take any point $latex P(x,y)$ on the parabola, draw $latex FP$, and draw $latex PQ$ perpendicular to line $latex l$ where Q is line l, then the distance between $latex F$ and $latex P$ and $latex P$ and $latex Q$ are equal.

parabola

 

Suppose that the coordinates of the focus is $latex (0,p)$ where $latex p > 0$, then the directrix is $latex y = -p$ (can you see why?).

From here we can see that $latex PQ = |y + p|$ and $latex PF = \sqrt{x^2 + (y – p)^2 }$. Since $latex PQ$ is equal to $latex PF$,

$latex |y + p| = \sqrt{x^2 + (y – p)^2 }$

Squaring both sides, we have

$latex (y + p)^2 = x^2 + (y-p)^2$
$latex y^2 + 2py + p^2 = x^2 + y^2 – 2py + y^2$
$latex 2py = x^2 – 2py$
$latex 4py = x^2$
$latex y = \frac{1}{4p}x^2$.

This is the equation of the parabola with focus $latex (0,p)$ and directrix $latex y = -p.$

Logarithm Base Changing Formula Proof

In a logarithmic expression, it is possible to change base using algebraic manipulation.  For example, we can change

$latex log_416$ to $latex \dfrac{\log_216}{\log_24}$.

In this post, we are going to prove why it is possible to do such algebraic manipulation. The change of base above can be generalized as

$latex \log_ab = \dfrac{\log_cb}{\log_ca}$.

Theorem

$latex \log_ab = \dfrac{log_cb}{\log_ca}$.

Proof

If we let $latex \log_ab = x$, then by definition, $latex a^x = b$.

Now, take the logarithm to the base $latex c$ of both sides. That is

$latex log_c a^x = \log_cb$.

Simplifying the exponent, we have

$latex x \log_ca = \log_cb$.

Now, since $latex a \neq 1$, $latex \log_ca \neq 0$.

Therefore,

$latex x = \dfrac{\log_cb}{\log_ca}$

Thus,

$latex \log_ab = \dfrac{log_cb}{\log_ca}$