MHB Proof of Equality for Odd Integers with Floor Function

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The discussion centers on proving the equality $$\left\lfloor{\frac{n}{2}}\right\rfloor= \left\lfloor{ \frac{n - 1}{ 2}}\right\rfloor$$ for odd integers n. The user starts with the expression for odd integers, $$n = 2k + 1$$, leading to $$\left\lfloor{\frac{2k + 1}{2}}\right\rfloor$$ which simplifies to $$\left\lfloor{k + \frac{1}{2}}\right\rfloor$$, resulting in $$k$$. They conclude that since $$k = \frac{n - 1}{2}$$, this demonstrates the equality holds true. The proof is deemed sufficient for the original question.
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I need to prove or disprove that

$$ \left\lfloor{\frac{n}{2}}\right\rfloor= \left\lfloor{ \frac{n - 1}{ 2}}\right\rfloor$$ where n is an odd integer.I start with something like,

$$\left\lfloor{\frac{2k + 1}{2}}\right\rfloor$$

and then

$$\left\lfloor{k + \frac{1}{2}}\right\rfloor$$ which equals $$k$$

But

$$ k = \frac{n - 1}{2}$$

So is that enough proof for the question? or is it wrong?
 
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tmt said:
I need to prove or disprove that

$$ \left\lfloor{\frac{n}{2}}\right\rfloor= \left\lfloor{ \frac{n - 1}{ 2}}\right\rfloor$$ where n is an odd integer.I start with something like,

$$\left\lfloor{\frac{2k + 1}{2}}\right\rfloor$$

and then

$$\left\lfloor{k + \frac{1}{2}}\right\rfloor$$ which equals $$k$$

But

$$ k = \frac{n - 1}{2}$$

So is that enough proof for the question? or is it wrong?
as n is odd a and n = 2k+1 so $ k = \frac{n - 1}{2}$ so it should be sufficient
 
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