# A Quick Proof: 08/11/17

'Sup, Nerds?

I am busily working on another post called *The Problem with Payloads*, within which I make that claim that for any positive numbers \(A\), \(B\), and \(C\), where \(A>B\), \begin{equation} \frac{A}{B} > \frac{A+C}{B+C} > 1. \label{081117frac} \end{equation}

Back during my days as a young math student, it was drilled into me that I should never declare such a mathematical statement without proof. Simultaneously, during my days as a young physics student, I cursed any author who uttered the phrase, "proof is left as an exercise for the reader."

"The burden of proof lies upon whoever makes a claim," was generally my retort, though the fact that I did not want extra work was a major contributing factor as well. Of course, I recognized the value of such exercises for building one's thinking skills, but I also believed that having a correct answer with which to compare my own was necessary for maximizing one's mental gains. (After all, if you get the wrong answer and have no way of knowing it, then what is the point of the problem?)

Therefore, while I strongly encourage all of my readers to try their own hands at proving these statements, I will always provide my own work for comparison. If we come up with the solution, fantastic! Either we are both correct, or we are both wrong. If our answers differ, then we have something to discuss.

So, without further ado, here we go!

My proof proceeds in two parts. First, I will prove that \(A/B > (A+C)/(B+C)\):\begin{equation*}

\begin{array}{r|l}

A > B\ &\text{Given}

\\[2ex]

A C > B C &\text{Multiplying both sides by $C$}

\\[2ex]

A B + A C > A B + B C &\text{Adding $A B$ to both sides}

\\[2ex]

A (B + C) > B (A + C) &\text{Factoring}

\\[2ex]

\displaystyle \frac{A}{B} > \frac{A + C}{B + C} &\text{Dividing both sides by $B(B+C)$}

\end{array}

\end{equation*}Then, I prove that \((A+C)/(B+C)>1\):\begin{equation*}

\begin{array}{r|l}

A > B\ &\text{Given}

\\[2ex]

A + C > B + C &\text{Adding $C$ to both sides}

\\[2ex]

\displaystyle \frac{A + C}{B + C} > 1 &\text{Dividing both sides by $B+C$}

\end{array}

\end{equation*}

Therefore,\begin{equation*} \frac{A}{B} > \frac{A+C}{B+C} > 1,\end{equation*} where \(A, B, C > 0\), and \(A>B\).

\(\therefore \text{QED}\)