Question
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Answer :
Word Count : 572
Let (R = \mathbb{Z}[\sqrt{2}]) and (M = {a + b\sqrt{2} \in R \mid 5 \mid a \text{ and } 5 \mid b}). We first check whether (M) is a subring of (R). For that, we need to verify if (M) is closed under addition, subtraction, and multiplication, and contains (0). 1. Closure under addition: Take (x = a_1 + b_1\sqrt{2} \in M) and (y = a_2 + b_2\sqrt{2} \in M). Then (a_1, a_2, b_1, b_2) are divisible by 5. [ x + y = (a_1 + a_2) + (b_1 + b_2)\sqrt{2} ] Since (5 \mid a_1, a_2) and (5 \mid b_1, b_2), we have (5 \mid (a_1 + a_2)) and (5 \mid (b_1 + _________ ____ _____ ________ _______ ________ _________ ______ _______.
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Let (R = \mathbb{Z}[\sqrt{2}]) and (M = {a + b\sqrt{2} \in R \mid 5 \mid a \text{ and } 5 \mid b}). We first check whether (M) is a subring of (R). For that, we need to verify if (M) is closed under addition, subtraction, and multiplication, and contains (0). 1. Closure under addition: Take (x = a_1 + b_1\sqrt{2} \in M) and (y = a_2 + b_2\sqrt{2} \in M). Then (a_1, a_2, b_1, b_2) are divisible by 5. [ x + y = (a_1 + a_2) + (b_1 + b_2)\sqrt{2} ] Since (5 \mid a_1, a_2) and (5 \mid b_1, b_2), we have (5 \mid (a_1 + a_2)) and (5 \mid (b_1 + _________ ____ _____ ________ _______ ________ _________ ______ _______.
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