Question

P rove that all real of the form a + b 2 , a; b ∈ Z form s a ring.

Answer 1

Given, $a + b\sqrt2 , a; b ∈ Z$

Proof:

A ring is a set R equipped with two binary operations + (addition) and ⋅ (multiplication) satisfying the following three sets of axioms, called the ring axioms

1.R is an abelian group under addition, meaning that:

$(a+b)+c=a+(b+c)$

we have:

$(a_1+b_1\sqrt2+a_2+b_2\sqrt2)+a_3+b_3\sqrt2=a_1+b_1\sqrt2+(a_2+b_2\sqrt2+a_3+b_3\sqrt2)$

$a+b=b+a$

we have:

$a_1+b_1\sqrt2+a_2+b_2\sqrt2=a_2+b_2\sqrt2+a_1+b_1\sqrt2$

There is an element 0 in R such that a + 0 = a

we have:

element 0 in R: $a=b=0$

For each a in R there exists −a in R such that a + (−a) = 0

we have:

$a+b\sqrt2+(-a-b\sqrt2)=0$

2.R is a monoid under multiplication, meaning that:

$(a ⋅ b) ⋅ c = a ⋅ (b ⋅ c)$we have:

$((a_1+b_1\sqrt2)(a_2+b_2\sqrt2))(a_3+b_3\sqrt2)=(a_1+b_1\sqrt2)((a_2+b_2\sqrt2)(a_3+b_3\sqrt2))$

There is an element 1 in R such that a ⋅ 1 = a and 1 ⋅ a = a

we have:

element 1 in R: $a=1,b=0$

3.Multiplication is distributive with respect to addition, meaning that:

a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

we have:

$(a_1+b_1\sqrt2)(a_2+b_2\sqrt2+a_3+b_3\sqrt2)=(a_1+b_1\sqrt2)(a_2+b_2\sqrt2)+$

$+(a_1+b_1\sqrt2)(a_3+b_3\sqrt2)$

(b + c) ⋅ a = (b ⋅ a) + (c ⋅ a)

we have:

$(a_2+b_2\sqrt2+a_3+b_3\sqrt2)(a_1+b_1\sqrt2)=(a_1+b_1\sqrt2)(a_2+b_2\sqrt2)+$

$+(a_1+b_1\sqrt2)(a_3+b_3\sqrt2)$