Bit of a dumb question

[cplangew]

Hey,

I was reviewing my calculus and came across an integration problem. I could do the integration but there was one part that I did not understand:

x / (1+x) = 1 - 1 / (1+x).

I've looked around the web, through my notes, etc, and I've seen it used, however, I cannot seem to grasp the intuition of this.

Any help with understand what is happening algebraically would be greatly appreciated.

 

[soroban]

Hello, cplangew!

There was one part that I did not understand: .\(\displaystyle \dfrac{x}{1+x} \:=\:1 - \dfrac{1}{1+x}\)

You can do long division on: .\(\displaystyle \dfrac{x}{x+1}\)


. . \(\displaystyle \begin{array}{cccccc} &&&& 1 \\ && --&--&-- \\ x+1 & | & x \\ &&x&+&1 \\ &&--&--&-- \\ &&& -&1\end{array}\)


Therefore: .\(\displaystyle \dfrac{x}{x+1} \:=\:1 - \dfrac{1}{x+1}\)

Or you can do this:

. . \(\displaystyle \displaystyle \frac{x}{1+x} \;=\;\frac{x \color{red}{+ 1 - 1}}{1+x} \;=\;\frac{1+x}{1+x} - \frac{1}{1+x}\)

 

[lookagain]

The following position of the "1" in the quotient is more standard, and the inclusion
of the "+ 0" in the quotient makes it clearer where the remainder of -1 comes from.


You can do long division on: .\(\displaystyle \dfrac{x}{x+1} \ \ \ or \ \ \ \dfrac{x + 0}{x+1}.\)


\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ ------\)
\(\displaystyle x + 1 \ \ \ \)|\(\displaystyle \ \ \ \ x \ \ \ + \ \ \ 0\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ + \ \ \ 1\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -----\)



You can think of the signs changing in the last row and then adding the last two rows:



\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ -------\)
\(\displaystyle x + 1 \ \ \ \)|\(\displaystyle \ \ \ \ \ \ \ x \ \ \ + \ \ \ 0\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -x \ \ \ - \ \ \ 1\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ------\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \ \ \ 1\)




.\(\displaystyle \dfrac{x}{x+1} \ \ = \ \ 1 \ - \ \dfrac{1}{x+1}\)

 

[Dale10101]

Reverse simplification

Hey,

I was reviewing my calculus and came across an integration problem. I could do the integration but there was one part that I did not understand:

x / (1+x) = 1 - 1 / (1+x).

I've looked around the web, through my notes, etc, and I've seen it used, however, I cannot seem to grasp the intuition of this.

Any help with understand what is happening algebraically would be greatly appreciated.

99% of math training involves simplifying expressions but in this case you are going form a simple to a less simple expression, so to speak. I remember running into this issue when proving basic limit problems.

To see what is happening take the right side of your statement and simplify it to the left side without leaving out any steps, pretty easy for you I bet, i.e simplify 1- 1/(x+1) and ponder the reverse procedure ... and Bob's your Uncle.

hint: (1+x)-1/(1+x) = x/(1+x)