Show that the set of all real numbers of the form \(a+b \sqrt{2}\), where \(a\) and \(b\) are ratioral numbers, is a ficld If \(a\) and \(b\) are restncted to the integers show that this set is a ring, but 15 not a field

We consider the set \(F = \{a+b\sqrt{2} : a,b \in Q\}\) where Q is the set of all rational numbers. We need to prove it is a field by verifying below properties1. Addition and multiplication are closed in F. Let's take two elements \(x = a + b\sqrt{2}\) and \(y = c + d\sqrt{2}\). Their sum \(x + y = (a + c) + (b + d)\sqrt{2}\) and their product \(xy = (ac + 2bd) + (ad + bc)\sqrt{2}\) are both in F.2. The set F has an additive identity '0' and each element in F has an additive inverse. For any element in F, \(a + b\sqrt{2}\), the additive inverse is \(-a - b\sqrt{2}\)3. The set F has a multiplicative identity '1', and element in F except '0' has a multiplicative inverse. Except when \(a = b = 0\), which gives '0', there exists a multiplicative inverse given by \(\frac{a - b\sqrt{2}}{a^2 - 2b^2}\) which belongs to F.4. The set F is associative and commutative under addition and multiplication.

Now we consider the set \(R = \{a+b\sqrt{2} : a,b \in Z\}\) where Z is the set of all integers. We follow similar steps as in Step 1, however in proving R to be a ring we are not obligated to show the existence of multiplicative inverses. By reusing and rechecking the proofs in step 1: 1. R is closed under addition and multiplication. 2. R has additive identity '0' and each element in R has an additive inverse.3. R has a multiplicative identity '1'.4. The set R is associative and commutative under addition and multiplication, and multiplication is distributive over addition.However, unlike in Step 1, each non-zero element of R does not necessarily have a multiplicative inverse in R. Therefore, R is a ring but not a field.