# For all real numbers $x$, prove $\lfloor x - 2\rfloor = \lfloor x\rfloor - 2$ [closed]

Prove the following statement:

For all real numbers $x$, $\lfloor x - 2\rfloor = \lfloor x\rfloor - 2$

I'd appreciate some help with this.

All I know is that the floor function $n$ implies : $n \leq x < n+1$

## closed as off-topic by user296602, JonMark Perry, Claude Leibovici, user26857, heropupFeb 29 '16 at 16:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, JonMark Perry, Claude Leibovici, user26857, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.

• Perhaps you meant $n\leq x<n+1$, not $x+1$. The question you're asking is a direct application of the definition. – Michael Burr Feb 29 '16 at 3:32
• Welcome to MSE! Can you share what you've tried, and what you're having trouble with? For example: What happens if you write $x = \lfloor x \rfloor + \epsilon$ with $0 \le \epsilon < 1$? Then what is $\lfloor x - 2\rfloor$? – user296602 Feb 29 '16 at 3:34
• Perhaps I don't understand math exchange well but why does a person need to nessecarily have attempted the question very far in depth? If the person has no idea what to do, then how are they supposed to get help from us? If a person asks an indefinite integration question but has no knowledge on how to take integrals, then why is right to turn them away? This question seems like a great way to showcase how different shifting translations can result in the same graphs of floor. – The Great Duck Mar 4 '16 at 5:03
• – user228113 Mar 10 '16 at 9:31

## 1 Answer

Let $x$ be a real number. Then by the well-ordering principle there exists a unique $n\in\mathbb{Z}$ such that $n\leq x<n+1$.

This implies that $\lfloor x\rfloor = n$ by definition of the floor function.

From here, you need to come up with an argument for why $\lfloor x-2\rfloor = n-2$.

Does it follow from the above that $n-2\leq x-2$?

Does it follow from the above that there is no integer larger than $n-2$, lets call it $m$, such that $n-2<m\leq x-2$?

• Actually, a better argument is that floor removes the fractional portion of its argument. -2 has no fraction portion so it can be removed with ease. In fact, this is true of any integer. Another way to think of it is as a graphical translation. Both shifting floor up and shifting floor left have the same results for integers. – The Great Duck Mar 4 '16 at 4:59