Question: Is f: Z2 -> Z2 with f(x) = -x a homomorphism?

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The discussion centers on whether the function f: Z2 -> Z2 defined by f(x) = -x is a homomorphism. The initial analysis confirms that f: Z -> Z is a homomorphism, as it satisfies the necessary properties. However, there is uncertainty regarding f: Z2 -> Z2, with doubts raised about its homomorphic nature. The conversation highlights confusion over the number of proofs presented and the clarity of the arguments. Ultimately, the conclusion remains that both cases were analyzed but the second case requires further clarification.
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(A) f: Z -> Z with a function defined as f(x)=-x

1.) f(0)=0
2.) f(a)-f(b)=-a+b=-a+b=f(a-b)
3.) -f(a)=a=a=f(-a)

Conclusion: a homomorphism

(B) f: Z2 -> Z2 with a function defined as f(x)=-x
This one I'm not sure, I doubt it is a homomorphism though.

f: g -> g
 
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Z_s? s equal what? s=2 certainly is a homomorphism. s>2 may cause problems. But not with f(0).
 


Sorry, I didn't even realize all the things i left off. But anyway, are my three proofs correct?
 


What three proofs? I thought there were maximum two problems here. f(x)=-x for i) Z->Z and ii) Z_2->Z_s. Whatever s is. Homomorphism, yes or no. You are being super unclear.
 


Wow, sorry. I just transposed everything from my notes quickly. The S was suppose to be a 2, which I fixed. I also only did put up two proofs just to see if Z -> Z and Z2 -> Z2 were homomorphisms. I arrived at the same conclusion for both. I actually didn't type out the third one, I'm not sure why.
 
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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