No, because \(\displaystyle \frac{3+x}{3} = \frac{3}{3}+\frac{x}{3} = 1 + \frac{x}{3}\).Can \(\displaystyle \frac{3+x}{3}\) be simplified to \(\displaystyle \frac{x}{3}\)

You can't just get rid of the 3 in the numerator.

Hope that helps

Mathemagister

Sure...but your final solution will be incorrect (Giggle)Can \(\displaystyle \frac{3+x}{3}\) be simplified to \(\displaystyle \frac{x}{3}\)

If x approaches infinity,Can \(\displaystyle \frac{3+x}{3}\) be simplified to \(\displaystyle \frac{x}{3}\)

then \(\displaystyle \frac{x+3}{3}\) will be indistinguishable from \(\displaystyle \frac{x}{3}\)

in other words x has to be so large that x is "no different" from x+3

Along these lines we would write that, for "large" x (in fact, for x with "large" absolute value),If x approaches infinity,

then \(\displaystyle \frac{x+3}{3}\) will be indistinguishable from \(\displaystyle \frac{x}{3}\)

in other words x has to be so large that x is "no different" from x+3

\(\displaystyle \frac{x+3}{3} \approx \frac{x}{3}\)

If this is part of some larger, more complicated equation, it may be useful to replace the original expression with an approximation. This practice is often done when modeling actual data in the sciences.

A more mathematically tight way of writing that would beAlong these lines we would write that, for "large" x (in fact, for x with "large" absolute value),

\(\displaystyle \frac{x+3}{3} \approx \frac{x}{3}\)

If this is part of some larger, more complicated equation, it may be useful to replace the original expression with an approximation. This practice is often done when modeling actual data in the sciences.

\(\displaystyle \lim_{x\rightarrow \pm \infty} \frac{3+x}{3} = \lim_{x\rightarrow \pm \infty} \frac{x}{3}\)

Let x = 1 in your proposed simplification. Does 4/3 = 1/3?Can \(\displaystyle \frac{3+x}{3}\) be simplified to \(\displaystyle \frac{x}{3}\)

Haha, I thought the amount of discussion generated by the original post was pretty silly too, but Archie Meade's post seemed like an interesting spin to me (like in those lateral thinking problems that I also often find silly, but can have interesting solutions), which I thought was worth a little extra commentary, and apparently mathemagister felt a little more extra commentary was appropriate in response to my post, which contributed to this thread being ridiculously long. Next we'll be discussing how the proposed simplification relates to the Riemann hypothesis and P vs NP. (Rofl)Thanks Mr F.

WHY spend time (even 1 second) in trying to somehow reason out

that (a + b) / b = a / b ....

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