Answer to Question #135827 in Abstract Algebra for YASHU

Question #135827
Show that the set of all real numbers of the form a+b√2 where 'a' and 'b' are integrals forms an integral domain
1
Expert's answer
2020-09-29T19:03:19-0400

Denote by "K" the set of all real numbers of the form "a+b\\sqrt{2}" where "a" and "b" are integers, that is

"K=\\{ a+b\\sqrt{2}\\ |\\ a,b\\in\\mathbb Z\\}."

To prove that "(K,+,\\cdot)" is a ring it is sufficient to prove that "(K,+,\\cdot)" is a subring of the ring "(\\mathbb R,+,\\cdot)" of real numbers.

Let "a+b\\sqrt{2}, c+d\\sqrt{2}\\in K", where "a,b,c,d\\in\\mathbb Z." Then "(a+b\\sqrt{2})+ (c+d\\sqrt{2})=(a+c)+(b+d)\\sqrt{2}\\in K", since "a+c,b+d\\in\\mathbb Z,"

"(a+b\\sqrt{2})- (c+d\\sqrt{2})=(a-c)+(b-d)\\sqrt{2}\\in K", since "a-c,b-d\\in\\mathbb Z,"

"(a+b\\sqrt{2})\\cdot (c+d\\sqrt{2})=(ac+2bd)+(ad+bc)\\sqrt{2}\\in K", since "ac+2bd, ad+bc\\in\\mathbb Z."

Therefore, "(K,+,\\cdot)" is a subring of the ring "(\\mathbb R,+,\\cdot)", that is "(K,+,\\cdot)" a ring.

Since "(\\mathbb R,+,\\cdot)" is a commutative ring, "(K,+,\\cdot)" is a commutative ring as well. The number "1=1+0\\sqrt{2}" is the identity of the ring "(K,+,\\cdot)". It follows that "(K,+,\\cdot)" is a commutative ring with identity. Since for any real numbers the equality "\\alpha\\cdot\\beta=0" implies that "\\alpha=0" or "\\beta=0," in particular, for "a+b\\sqrt{2}, c+d\\sqrt{2}\\in K", the equality "(a+b\\sqrt{2})(c+d\\sqrt{2})=0" implies that "a+b\\sqrt{2}=0" or "c+d\\sqrt{2}=0."

Therefore, "(K,+,\\cdot)" is an integral domain.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS