Every ring with unity has at least two units

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Every ring with unity does not necessarily have at least two units. The discussion highlights that while unity (1) and its additive inverse (-1) can both be units, they may not be distinct in certain rings, such as when 1 equals -1. The zero ring is mentioned as an example that lacks any units, prompting the need for clarification on what constitutes a unit. The conversation suggests that there are non-trivial rings where this condition holds, indicating that the initial assumption of the statement being true is incorrect. Understanding the properties of units in different rings is essential for accurate conclusions.
Mr Davis 97
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Homework Statement


T/F: Every ring with unity has at least two units

Homework Equations

The Attempt at a Solution


I thought that the answer was true, because if a ring ##R## has unity ##1##, then ##1 \cdot 1 = 1## and ##(-1) \cdot (-1) = 1##. Where am I going wrong?
 
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Mr Davis 97 said:
I thought that the answer was true, because if a ring ##R## has unity ##1##, then ##1 \cdot 1 = 1## and ##(-1) \cdot (-1) = 1##. Where am I going wrong?

The elements denoted by "##a##" and "##-a##" are not necessarily distinct from each other. Can you think of a ring where 1 = -1 ?
 
Stephen Tashi said:
The elements denoted by "##a##" and "##-a##" are not necessarily distinct from each other. Can you think of a ring where 1 = -1 ?
I can only think of the zero ring as an example. Are there non-trivial examples?
 
Mr Davis 97 said:
I can only think of the zero ring as an example. Are there non-trivial examples?
Many. The zero ring hasn't any units at all. So what do you need also to speak of units and what is the smallest example? Based on this, there are plenty of examples.
 
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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