MHB What is [2x] Notation? Continuity Question

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The discussion clarifies that the notation [2x] refers to the floor function, which rounds down to the nearest integer below the value of 2x. Participants confirm that for values like 1.2 or 1.99, the floor function results in 1. The question also revolves around determining the continuity of the function defined in the problem. The floor function's behavior is essential for understanding the function's continuity in the specified intervals. Overall, the floor function notation is key to solving the continuity question.
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The question asks me to determine whether the function is continuous?
f(x)=1, x =0,1
f(x)=x+[2x], 0<x<1
what is this [2x]? I cannot find it in the textbook and during lecture we had no information given about this.View attachment 6413
 

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FallArk said:
The question asks me to determine whether the function is continuous?
f(x)=1, x =0,1
f(x)=x+[2x], 0<x<1
what is this [2x]? I cannot find it in the textbook and during lecture we had no information given about this.

Hi FallArk!

The $[x]$ notation indicates the floor function.
That is, rounding down to the nearest integer below $x$.
Btw, I prefer the notation $\lfloor x \rfloor$, which is more intuitive.
 
I like Serena said:
Hi FallArk!

The $[x]$ notation indicates the floor function.
That is, rounding down to the nearest integer below $x$.
Btw, I prefer the notation $\lfloor x \rfloor$, which is more intuitive.

Then if i get 1.2, the result would be 1?
 
FallArk said:
Then if i get 1.2, the result would be 1?

Yes.
$$[1] = [1.2] = [1.99] = 1$$
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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