• One of the first examples of a noncommutative division ring was found by the Irish mathematician William Hamilton in the 1840's. His example, the ring of real quaternions H(R) consists of all...

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• One of the first examples of a noncommutative division ring was found by the Irish mathematician William Hamilton in the 1840's. His example, the ring of real quaternions H(R) consists of all elements {a + bi + cj + dk I a, b, c, d E R} for which j, k satisfy the equations i2= j2 = k2 = -1, i • j = k = -j • i, j • k = i = -k • j, k • i = j = -i • k. (Note, these are the same i, j, k that define the complex cube using the right hand rule.)
We add two quaternions componentwise using real arithmetic and multiply using distributive laws (there would be 16 terms in the distribution.) Show that H(R) is a division ring. That is, show that H(R) is a ring with unity for which every element is a unit. Then show that elements of H(R) do not commute. Hence we have a division ring that is not a field.
For the discussion prompt, attempt a proof without referring to your peer's responses. Once your proof is posted, respond to peers with an analysis and critique of their proof. Be constructive in your responses.
The Gaussian integers, together with the ordinary addition and multiplication of complex numbers, form an integral domain written as 71[i]. This interesting integral domain has many applications and quite a few unsolved problems. For this discussion, research Gaussian integers and report on one of the unsolved problems associated with Gaussian integers.



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Answered Same DayDec 20, 2021

Answer To: • One of the first examples of a noncommutative division ring was found by the Irish mathematician...

David answered on Dec 20 2021
132 Votes
1. Following we will try to prove the condition which are required for Z ×R
to be a Ring:
(a) Clos
ed under addition and Multiplication: As +, . is binary operation
for Z, and ./, ◦ is binary operation for R, Hence
(m+ n, a ./ b) ∈ Z×R
If we assume m ◦ b, n ◦ a ∈ R, then m ◦ b ./ n ◦ a ∈ R. Hence
m ◦ b ./ n ◦ a+ a ◦ b ∈ R. This gives
(m.n,m ◦ b ./ n ◦ a+ a ◦ b) ∈ Z×R
Hence operation ⊕ and � are binary operation in Z×R.
(b) Associative of addition and Multiplication: For m,n, p ∈ Z and
a, b, c ∈ R, we have: m + (n + p) = (m + n) + p and a ./ (b ./
c) = (a ./ b) ./ c, we have
(m, a)⊕ ((n, b)⊕ (p, c)) = ((m, a)⊕ (n, b))⊕ (p, c)
That is ⊕ is associative. Same way we have Now for �, we have
(m, a)� ((n, b)� (p, c)) = (m, a)� (n.p, n ◦ c ./ p ◦ b ./ b ◦ c)
= (m.(n.p),m ◦ (n ◦ c ./ p ◦ b ./ b ◦...
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