Proof that $\frac {(2+t)}{(1+t)} = 1 +\frac{ 1}{(1+t)}$?

Question

I might just be out of my mind and can't see a simple solution. I'm just running into a basic simplification problem based off of an instructor's solution.

For more context, this is part of a parametric equation problem, where $x(t) = \frac{(2+t)}{(1+t)}$ and $y(t) = \frac{1}{(1+t)^3} +1 $. The solution from my instructor is that $y = (x-1)^3 + 1$, with the $x$ equation simplifying to $1 + \frac{1}{(1+t)}$. I just don't understand that one step. Am I missing something?

Also, I'm extremely new to the forums. So, if I messed up formatting or purely just content, I am deeply sorry and would love to know what to do to fix it or what to do next time.

Thanks for any help!

Answer 1

$$\frac{2+t}{1+t}=\frac{1+1+t}{1+t}=\frac{1}{1+t}+\frac{1+t}{1+t}=\frac{1}{1+t}+1$$

Answer 2

A good way to prove that two things are equal is to prove that their difference is zero. In this approach, the task of a "convenient manipulation" is replaced by a "pure simplification" and thus the work becomes easier: $$\frac{2+t}{1+t} - \left(1 + \frac{1}{1+t}\right)=\frac{2+t}{1+t}- \frac{1}{1+t}-1=\frac{2+t-1}{1+t}-1=1-1=0.$$

Note that in the Zachary Selk's answer (that proves the equality directly instead of computing the difference) there are two convenient manipulations:

• write $2$ as $1+1$ (instead of $3-1$, for example).
• write $\frac{1+1+t}{1+t}$ as $\frac{1}{1+t}+\frac{1+t}{1+t}$ (instead of $\frac{1+1}{1+t}+\frac{t}{1+t}$, for example).

To consider the difference is more effective especially if we don't know a priori that the equality is valid. However, a "direct proof" usually is more elegant.

Answer 3

We have $$(1+t)\left(1+\frac{1}{1+t}\right) = (1+t) + 1 = 2+t.$$ Now divide both sides by $1+t.$