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I ask because it occurred to me that in certain cases there are 2 or more ways of representing an expression/equation each of which seems equally 'simple' to me (though I could be wrong). This can be problematic because these different representations don't look on the surface very similar.

For example: $\frac{(e^t)/2-2}{(e^t)/2-1}=\frac{4(e^{-t})-1}{2(e^{-t})-1}$ In each case there are the same number of terms in numerator & denominator...

In cases like this are there best practices for representing an expression?

profPlum
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  • The numerator and denominator should have no common factors . – Vivaan Daga Jun 29 '22 at 15:08
  • I do not think there is a rigorous definition for this. Which one is simpler: $x\sqrt x$ or $x^{\frac{3}{2}}$ or $\sqrt{x^3}$? – Vasili Jun 29 '22 at 15:22
  • I've converted your math to LaTeX/MathJax. If you hit the 'Edit' link, you can see what the appropriate MathJax looks like. From now on it's you're job to typeset your math in the way that is easier for others to read. – JonathanZ Jun 29 '22 at 15:46

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I guess it really depends on what you want to do with your expression. Take the following examples.

  1. For $n>1$ a natural number consider the expression $1/n(n-1)$, which looks simple enough (doesn't it?). We have the identity $$\frac{1}{n(n-1)} = \frac{1}{n-1}-\frac{1}{n}.$$ When you want to calculate the sum $$\sum_{n=2}^{\infty} \frac{1}{n(n-1)}$$ the first expression at first does not look so helpful: $$\sum_{n=2}^{\infty} \frac{1}{n(n-1)} = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\dots = \text{?}$$ However, the second one reveals a nice pattern: $$\sum_{n=2}^{\infty} \frac{1}{n(n-1)}= \sum_{n=2}^{\infty} \frac{1}{n-1}-\frac{1}{n} = 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots = 1.$$

  2. Suppose you want to find the solutions of the inequality $$\frac{1}{x-1}-\frac{1}{x} > 0, \quad \text{ for }x < 0.$$ As it is written, it is perhaps not obvious to understand the set of solutions, because we are adding a negative number to a positive one, so it would be about understanding when the positive one wins. But using the above identity we find the inequality $$\frac{1}{x(x-1)} > 0.$$ We know that $x$ is negative, so $1/x$ is also negative, hence by the product rule we need to have $$\frac{1}{x-1} < 0,$$ which is true when $x-1 < 0$, i.e. $x < 1$. We are asking $x < 0$, so all negative numbers are solutions.

Gibbs
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    Sometimes we need to un-simplify. I was helping someone with algebra. The topic was rationalizing a denominator. I wrote $\frac {1}{\sqrt 2,-1}=\frac {\sqrt 2,+1}{(\sqrt 2,-1)(\sqrt 2,+1)}$ and he immediately said the entire latter expression simplifies to $\frac {1}{\sqrt 2,-1}$, but then we're back where we began. But $part$ of the latter expression (the denominator) simplifies to $1$. – DanielWainfleet Jun 29 '22 at 16:19