How can I prove the continuity of the ceiling function at a non-integer value?

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The discussion focuses on proving the continuity of the ceiling function at non-integer values. The ceiling function, defined as f(x) = ceil(x), is continuous at non-integers because within a small interval around a non-integer 'a', the function remains constant. Participants highlight that for any ε>0, one can find a δ>0 such that |f(x) - f(a)| < ε, as f(x) does not change in that interval. The key insight is that since 'a' is not an integer, the values of f(x) will not change, leading to continuity. The thread emphasizes understanding the piecewise nature of the function and its graphical representation.
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Homework Statement



Let f: R -> Z be the ceiling function defined by f(x) = ceil(x). Give a ε-δ proof that if a is a real number that is not an integer, then f is continuous at a.


The Attempt at a Solution



I can prove that f(x) is not continuous at any integer. But i don't know how to prove this. I can do proofs for continuous functions, but I've never done one for a piece wise function. Any help would be awesome. Thanks.
 
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Think about what the graph of this function looks like -- essentially steps that are 1 unit wide. If a is not an integer, it shouldn't be too hard to find a number δ > 0, such that |f(x) - f(a)| < ε. In fact, if x is close enough to a, |f(x) - f(a)| will be 0.
 
If a is not an integer, then there exist \delta&gt; 0 such that every number in the interval from a- \delta to a+ \delta is not an integer. On that interval f(x) is a constant.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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