Let k be the set of all numbers which can be written in the form, where a, b are rational numbers. Show that k is a field.
To show that k is a field, we need to show that k satisfies all field axioms.
1. Closure under addition:
Let x, y be two elements in k, written as x = a/b and y = c/d, where a, b, c, d are rational numbers.
Then x + y = (a/b) + (c/d) = (ad + bc)/bd, where ad + bc and bd are both rational numbers.
Therefore, x + y is an element in k.
2. Closure under multiplication:
Let x, y be two elements in k, written as x = a/b and y = c/d, where a, b, c, d are rational numbers.
Then x * y = (a/b) * (c/d) = (ac)/(bd), where ac and bd are both rational numbers.
Therefore, x * y is an element in k.
3. Commutativity of addition:
For any x, y in k, x + y = (a/b) + (c/d) = (ad + bc)/bd and y + x = (c/d) + (a/b) = (ad + bc)/bd.
Since addition of rational numbers is commutative, we have x + y = y + x.
4. Commutativity of multiplication:
For any x, y in k, x * y = (a/b) * (c/d) = (ac)/(bd) and y * x = (c/d) * (a/b) = (ac)/(bd).
Since multiplication of rational numbers is commutative, we have x * y = y * x.
5. Associativity of addition:
For any x, y, z in k, (x + y) + z = ((a/b) + (c/d)) + (e/f) = ((ad + bc)/bd) + (e/f)
= ((ad + bc)f + bde)/(bdf).
On the other hand, x + (y + z) = (a/b) + ((c/d) + (e/f)) = (a/b) + ((cf + de)/df)
= ((adf + bcf) + bde)/(bdf).
Since addition of rational numbers is associative, we have (x + y) + z = x + (y + z).
6. Associativity of multiplication:
For any x, y, z in k, (x * y) * z = ((a/b) * (c/d)) * (e/f) = ((ac)/(bd)) * (e/f)
= (ace)/(bdf).
On the other hand, x * (y * z) = (a/b) * ((c/d) * (e/f)) = (a/b) * ((ce)/(df))
= (ace)/(bdf).
Since multiplication of rational numbers is associative, we have (x * y) * z = x * (y * z).
7. Additive identity:
The additive identity in k is 0, which is the rational number 0/1.
For any x in k, x + 0 = (a/b) + 0 = (a/b) + (0/1) = (a/b) * (1/1) + (0/1) * (b/b)
= (a/b) * (1/b) + (0/b) * (b/1) = ab/(bb) + 0/(b) = ab/(bb).
Since ab/(bb) is a rational number and b is non-zero, ab/(bb) is in k.
Therefore, 0 is the additive identity in k.
8. Multiplicative identity:
The multiplicative identity in k is 1, which is the rational number 1/1.
For any x in k, x * 1 = (a/b) * (1/1) = (a/b) * (b/b) = (ab)/(bb) = a/b.
Since a/b is a rational number, it is in k.
Therefore, 1 is the multiplicative identity in k.
9. Additive inverse:
For any x in k, the additive inverse of x is -x, which is the rational number -a/b.
x + (-x) = (a/b) + (-a/b) = (a/b) - (a/b) = (ab - ab)/(bb) = 0/1 = 0,
which is the additive identity in k.
10. Multiplicative inverse:
For any non-zero x in k, the multiplicative inverse of x is 1/x, which is the rational number (b/a).
x * (1/x) = (a/b) * (b/a) = (ab)/(ba) = 1/1 = 1,
which is the multiplicative identity in k.
Since k satisfies all field axioms, k is a field.
To show that k is a field, we need to demonstrate that it satisfies the properties of a field.
1. Closure under addition: Let x and y be elements in k, where x = a/b and y = c/d, with a, b, c, and d being rational numbers. The sum x + y can be written as (ad + bc)/(bd). Since both ad + bc and bd are rational numbers, we can conclude that x + y is also in k.
2. Closure under multiplication: Let x and y be elements in k, where x = a/b and y = c/d, with a, b, c, and d being rational numbers. The product xy can be written as (ac)/(bd). Since both ac and bd are rational numbers, we can conclude that xy is also in k.
3. Associativity of addition: For all x, y, and z in k, we have (x + y) + z = x + (y + z). This property holds true because addition of rational numbers is associative.
4. Associativity of multiplication: For all x, y, and z in k, we have (xy)z = x(yz). This property holds true because multiplication of rational numbers is associative.
5. Existence of additive identity: There exists an element 0 in k such that for all x in k, x + 0 = x. The additive identity is 0/1, as (a/b) + (0/1) = (a/b) = (a/b) + 1.
6. Existence of multiplicative identity: There exists an element 1 in k such that for all x in k, x * 1 = x. The multiplicative identity is 1/1, as (a/b) * (1/1) = (a/b) = (a/b) * 1.
7. Existence of additive inverse: For every x in k, there exists an element -x in k such that x + (-x) = 0. The additive inverse of a/b is -a/b, as (a/b) + (-a/b) = (a/b) - (a/b) = 0.
8. Existence of multiplicative inverse: For every non-zero x in k, there exists an element 1/x in k such that x * (1/x) = 1. The multiplicative inverse of a/b is b/a, as (a/b) * (b/a) = (ab)/(ba) = 1.
Therefore, since k satisfies all the properties of a field, we can conclude that k is a field.