## Thursday, September 24, 2009

### The Conjugate of a Complex Number

Definition 1: conjugate of complex number

Let z=(a,b) be a complex number. Then z = (a,-b)

For a review of complex numbers, see here.

Theorem 1: x is a real number if and only x = x

Proof:

(1) Assume that x is a real number such that x = (x,0)

(2) x = (x,-0) = (x,0) = x

(3) Assume that x = x

(4) Let x = (a,b) so that we have:

(a,b) = (a,-b)

(5) But by the definition of equality for complex numbers (see definition 4, here), this is only true if a=a and b=-b

(6) But b=-b only if b+b=2b=0 so b = 0.

(7) So x must be a real number.

QED

Theorem 2:

Let ηi be an nth root of unity that is not real.

Then:

ηi = η-i

Proof:

(1) Any root of unity has the following form (see Corollary 1.1, here):

(2) So, there exists an integer k such that:

ηi = cos [(2kπ)/n] + isin[(2kπ)/n]

(3) The conjugate of this value is (see Definition 1 above):

cos[(2kπ)/n] - isin[(2kπ)/n]
(4) We note that:

(5) Using the well known cos2(x) + sin2(x) = 1 [see Corollary 2, here], we get:

1/(cos[(2kπ)/n] - isin[(2kπ)/n]) = cos [(2kπ)/n] + isin[(2kπ)/n]

(6) Which shows that:

cos[(2kπ)/n] - isin[(2kπ)/n] = 1/ηi = η-i

QED

Lemma 3: (a + b) = a + b

Proof:

(1) Let a = s + ti

(2) Let b = u + vi

(3) a + b = (s + u) + (t+v)i

(4) a + b = (s + u) - (t+v)i

(5) a + b = s - ti + u - vi = (s + u) - (t+v)i

QED

Lemma 4: (ab) = a * b

Proof:

(1) Let a = s + ti

(2) Let b = u + vi

(3) a * b = (s*u - t*v) + (s*v + u*t)i

(4) a * b = (s*u - t*v) - (s*v+u*t)i

(5) a * b = (s - ti)*(u - vi) = (s*u - t*v) - (s*v + u*t)i

QED

Theorem 5:

Let f(x) be a polynomial with real coefficients.

if r is a root of f(x), then r is also a root

Proof:

(1) Let f(x) = a0xn + a1xn-1 + ... + an

(2) Since r is a root, we have:

a0rn + a1rn-1 + ... + an = 0

(3) But then taking the complex conjugate of both sides, we get (using Theorem 1 above as well as Lemma 3 and Lemma 4 above):

a0rn + a1rn-1 + ... + an = 0

(4) Which shows that f(r) = 0.

QED

## Tuesday, September 22, 2009

### Complex Conjugates and Properties of Complex Numbers

The content in today's blog is taken from Bruce E. Meserve's Fundamental Concepts of Algebra.

The presentation builds on a previous blog I did on the definitions needed to build the complex numbers from the real numbers.

Definition 1: Complex Conjugate

For any complex number a+bi (see Definition 6, here), the complex conjugate is the form a-bi.

In other words, for the complex number (a,b), its complex conjugate is (a,-b). The complex conjugate of (a,-b) is likewise (a,b).

Definition 2: Norm of complex number: n(z)

The norm of a complex number which is represented as n(z) is the product of a complex number with its conjugate.

This definition is demonstrated in the following lemma:

Lemma 1: For any complex number (a,b), its norm is a real number: (a2 + b2,0)

Proof:

(1) Let (a,b) be any complex number.

(2) It's complex conjugate is (a,-b) [see Definition 1 above]

(3) Using the definition for multiplication of complex numbers (see Definition 3, here) and the definition for norms (see Definition 2 above):

It's norm is (a,b)*(a,-b) = (a*a - (b*-b),a*(-b) + (b*a)) = (a2 + b2,0)

(4) We can see that this is a real number (see Theorem 15, here)

QED

Definition 3: Absolute value of a complex number (the modulus)

The absolute value of a complex number is the nonnegative square root of the norm so that:

if z = (a,b), then abs(z) = √a2 + b2

Definition 4: Division of complex numbers

(a,b) / (c,d) = (p,q) if and only if (a,b) = (c,d)*(p,q)

That this definition is well-defined is established in the next theorem.

Theorem 2: For the set of complex numbers, division by an nonzero number is well-defined.

Proof:

(1) Let (a,b), (c,d) be two complex numbers.

(2) Using the Definition of Multiplication (see Definition 3, here):

(c,d)*(p,q) = (cp - dq,cq + dp)

(3) So, if (a,b) = (c,d)*(p,q), then we have (see Definition 4, here):

a = cp - dq b = cq + dp

(4) Solving the first equation in terms of p and the second equation in terms of q gives us:

p = (a + dq)/c q = (b - dp)/c

(5) Combining the equations for p, we have:

(6) Now, rearranging the equation gives us:

And solving for p gives us:

which results in:

(7) Combining the equations for q, we have:

(8) Now, rearranging the equation gives us:

And solving for q gives us:

which results in:

(9) By assumption (c,d) is not nonzero so c2 + d2 is nonzero

(8) Since a,b,c,d are real numbers and the operations on real numbers are well-defined, it follows that p,q are well-defined and therefore division of (a,b) by (c,d) is well-defined.

QED

Theorem 3: The norm of a product is equal to the product of the norms.

Proof:

(1) Using Lemma 1 above, the norm for (a,b) is (a2 +b2,0)

(2) Likewise, the norm for (c,d) is (c2 + d2,0)

(3) Using the definition of multiplication for complex numbers (see Definition 3, here),

(a,b)*(c,d) = (a*c -b*d,a*d + b*c)

(4) Norm(a*c - b*d,a*d + b*c) is ([ac-bd]2 + [ad+bc]2,0)

(5) Finally:

(ac - bd)2 + (ad + bc)2 = a2c2 - 2abcd + b2d2 + a2d2 + 2abcd + b2c2 =

= a2c2 + b2d2 + a2d2 + b2c2 = (a2 + b2)(c2 + d2)

which shows that:

Norm[(a,b)*(c,d)] = Norm(a,b)*Norm(c,d)

QED

Theorem 4:

The absolute value of a sum of complex numbers is less than or equal to the sum of the absolute values:

abs(z1 + z2) ≤ abs(z1) + abs(z2)

Proof:

(1) Assume that abs(z1 + z2) is greater than abs(z1) + abs(z2)

(2) Let:

z1 = a + bi

z2 = c + di

(3) Using the definition of absolute values for complex numbers (see Definition 3 above):

(a + c)2 + (b + d)2 is greater than a2 + b2 + √c2 + d2

(4) Squaring both sides gives us:

(a + c)2 + (b+d)2 is greater than a2 + b2 + 2√(a2 + b2)(c2 + d2) + c2 + d2

So that:

a2 + c2 + b2 + d2 + 2ac + 2bd is greater than a2 + b2 + c2 + d2 + 2√(a2 + b2)(c2 + d2)

So that:

ac + bd is greater than (a2 + b2)(c2 + d2)

(5) Squaring both sides gives us:

a2c2 + b2d2 + 2abcd is greater than a2c2 + b2d2 + a2d2 + b2c2

So that:

0 is greater than a2d2 + b2c2 +-2abcd

So that:

0 is greater than (bc - ad)2

(6) But this is impossible since (bc -ad)2 ≥ 0.

(7) So, we reject our assumption in step #1.

QED

References
• Bruce E. Meserve, , Dover Books, 1981.

## Monday, September 21, 2009

### The Set of Complex Numbers

In today's blog I go over the definition of complex numbers and show how this definition can be used to prove that the set of complex numbers forms a field.

Definition 1: Complex Number

A complex number is an ordered pair of real numbers: (x,y) where x,y are real numbers.

Definition 2: Addition of Complex Numbers

(a,b) + (c,d) = (a+c,b+d)

Definition 3: Multiplication of Complex Numbers

(a,b)*(c,d) = (a*c - b*d,a*d + b*c)

Definition 4: Equality

(a,b) = (c,d) if and only if a = c and b = d

Lemma 1: The set of complex numbers is closed on addition

Proof:

This follows directly from the fact that the real numbers are closed on addition [see Lemma 1, here] and Definition 2 above.

QED

Lemma 2: The set of complex numbers is closed on multiplication.

Proof:

This follows directly form the fact that the real numbers are closed on multiplication [see Lemma 2, here] and Definition 3 above.

QED

Lemma 3: The set of complex numbers supports the commutative rule for addition

Proof:

(1) By Definition 2 above:

(a,b) + (c,d) = (a+c,b+d)

(2) Since the real numbers support the commutative rule for addition [see Lemma 3, here]:

(a+c,b+d) = (c+a,d+b) = (c,d) + (a,b)

QED

Lemma 4: The set of complex numbers supports the associative rule for addition

Proof:

(1) By Definition 2 above:

[(a,b) + (c,d)] + (e,f) = (a+c,b+d) + (e,f) = ([a+c]+e,[b+d]+f)

(2) Since the real numbers support the associative rule for addition [see Lemma 4, here]:

([a+c]+e,[b+d]+f) = (a+[c+e],b+[d+f]) = (a,b) + [(c,d) + (e,f)]

QED

Lemma 5: The set of complex numbers support the commutative rule for multiplication

Proof:

(1) By definition 3 above, we have:

(a,b)*(c,d) = (a*c - b*d,a*d + b*c)

(2) Since the real numbers support the commutative rule for multiplication (see Lemma 5, here):

(a*c - b*d,a*d + b*c) = (c*a - d*b,d*a + c*b)

(3) Since the real numbers support the commutative rule for addition (see Lemma 3, here):

(c*a - d*b,d*a + c*b) = (c*a - d*b,c*b + d*a)

(4) Using Definition 3 above again:

(c,d)*(a,b) = (c*a - d*b,c*b + d*a)

QED

Lemma 6: The set of complex numbers support the associative rule for multiplication

Proof:

By definition 3 above, we have:

[(a,b)*(c,d)]*(e,f) = ([a*c - b*d],[a*d + b*c])*(e,f) =

= ([a*c-b*d]*e - [a*d+b*c]*f,[a*c - b*d]*f + [a*d+b*c]*e) =

= (a*c*e - b*d*e -a*d*f + b*c*f, a*c*f - b*d*f + a*d*e + b*c*e) =

= (a*[c*e - d*f] - b*[c*f + d*e],a*[c*f + d*e] + b*[c*e - d*f]) =

= (a,b)*([c*e - d*f],[c*f + d*e]) = (a,b)*[(c,d)*(e,f)]

QED

Lemma 7: The set of complex numbers support the distributive rule

Proof:

(1) By Definition 2 above:

(a,b)*[(c,d) + (e,f)] = (a,b)*(c+e,d+f)

(2) By Definition 3 above:

(a,b)*(c+e,d+f) = (a*(c+e) - b*(d+f),a*(d+f) + b*(c+e)) =

= (a*c+a*e - b*d -b*f, a*d+a*f +b*c+b*e) =

= ([a*c - b*d] + [a*e - b*f],[a*d + b*c] + [a*f + b*e]) =

= (a*c-b*d,a*d + b*c) + (a*e-b*f,a*f + b*e) =

= (a,b)*(c,d) + (a,b)*(e,f)

QED

Lemma 8: The set of complex numbers have an additive identity (0,0)

Proof:

(a,b) + (0,0) = (a+0,b+0) = (a,b)

QED

Lemma 9: Every element of the set of complex numbers has an additive inverse (-a,-b)

Proof:

(a,b) + (-a,-b) = (a+-a,b+-b) = (0,0)

QED

Lemma 10: The set of complex numbers has a multiplicative identity (1,0)

Proof:

(a,b)*(1,0) = (a*1 - b*0, a*0 + b*1) = (a,b)

QED

Lemma 11: If (a,b) ≠ (0,0) then a2 + b2 ≠ 0

Proof:

(1) From Definition 4 above:

(a,b) ≠ (0,0) implies either a ≠ 0 or b ≠ 0.

(2) a≠ 0 → a2 is greater than 0

(3) b≠ 0 → b2 is greater than 0

(4) So a2 is 0 or positive and b2 is 0 or positive

(5) In all cases, a2 + b2 is positive since:

position + 0 = positive

0 + positive = positive

positive + positive = positive

QED

Lemma 12: Every nonzero element of the set of complex numbers has a multiplicative inverse

Proof:

(1) Let (a,b) be any nonzero element of the set of complex numbers

(2) Let c =1/(a2 + b2)

We know that a2 + b2 is nonzero from Lemma 11 above.

(3) The multiplicative inverse is (a*c,-b*c)

(4) And we see that:

(a,b)*(a*c,-b*c) = (a*a*c - b*(-b*c),a*(-b*c) + b*(a*c)) =

= (a*a*c +b*b*c,-a*b*c + a*b*c) = (a*a*c + b*b*c,0)

(5) a*a*c + b*b*c = a*a/(a*a + b*b) + b*b/(a*a + b*b) =

= (a*a + b*b)/(a*a + b*b) = 1

QED

Theorem 13: The complex numbers form a field

Proof:

(1) The complex numbers are closed on addition [see Lemma 1 above] and multiplication [see Lemma 2 above].

(2) The complex numbers support the commutative property of addition [see Lemma 3 above], the associative property of addition [see Lemma 4 above], the commutative property of multiplication [see Lemma 5 above], an associative property of multiplication [see Lemma 6 above], and a distributive property [see Lemma 7 above].

(3) The set of complex numbers has an additive identity property [see Lemma 8 above], an additive inverse property [see Lemma 9 above], a multiplicative identity property [see Lemma 10 above], and a multiplicative inverse property [see Lemma 12 above].

(4) From all these properties, the complex numbers form a field. [see Definition 3, here]

QED

Definition 5: i

i = (0,1)

Theorem 14: i2 = (-1,0)

Proof:

The result follows directly from Definition 3 above:

(0,1)*(0,1) = (0*0 - 1*1,0*1 - 1*0) = (-1,0)

QED

Theorem 15: All real numbers can be represented as complex numbers

(1) Any real number x can be represented as (x,0) [see definition 1 above]

(2) This correspondence holds over addition

x + y = z if and only if (x,0) + (y,0) = (x+y,0) = (z,0)

(3) This correspondence holds over multiplication

x*y = z if and only if (x,0)*(y,0) = (x*y - 0*0,x*0 + 0*y) = (x*y,0) = (z,0)

QED

Definition 6: a+bi

a+bi = (a,b)

References
• H.A. Thurston, , Dover Publications, 1967
• Bruce E. Meserve, , Dover Books, 1981.

### The Set of Real Numbers

In a previous blog, I showed how the Dedekind cut could be used to define the real numbers.

In today's blog, I will show that the real numbers form a field.

Lemma 1: The real numbers are closed on addition.

Proof:

(1) We can define the real numbers based on a Dedekind cut. [see Definition 2, here]

(2) Let x,y be the real numbers.

(3) From the definition of the Dedekind cut, x is the set of rational numbers that are less than x and y is the set of rational numbers that are less than y.

(4) x+y is defined as the set of rational numbers in x added to the set of rational numbers in y so that x+y is the set of all of possible sums.

(5) Since the rational numbers are closed on addition [see Lemma 2, here], it follows that x+y is also closed on addition.

QED

Lemma 2: The real numbers are closed on multiplication

Proof:

(1) We can define the real numbers based on a Dedekind cut. [see Definition 2, here]

(2) Let x,y be the real numbers.

(3) From the definition of the Dedekind cut, x is the set of rational numbers that are less than x and y is the set of rational numbers that are less than y.

(4) xy is defined as the set of rational numbers in x multiplied to the set of rational numbers in y so that xy is the set of all of possible products.

(5) Since the rational numbers are closed on multiplication [see Lemma 3, here], it follows that xy is also closed on multiplication.

QED

Lemma 3: The set of real numbers support the commutative rule for addition

Proof:

(1) By the definition of addition for real numbers, addition follows the properties of the set of rationals. [see Definition 5, here]

(2) So, the conclusion follows from the fact that the rational numbers support the commutative rule for addition. [see Lemma 4, here]

QED

Lemma 4: The set of real numbers support the associative rule for addition

Proof:

(1) By the definition of addition for real numbers, addition follows the properties of the set of rationals. [see Definition 5, here]

(2) So, the conclusion follows from the fact that the rational numbers support the associative rule for addition. [see Lemma 5, here]

QED

Lemma 5: The set of real numbers support the commutative rule for multiplication.

Proof:

(1) By the definition of multiplication for real numbers, multiplication follows the properties of the set of rationals. [see Definition 7, here]

(2) So, the conclusion follows from the fact that the rational numbers support the commutative rule for multiplication. [see Lemma 11, here]

QED

Lemma 6: The set of real numbers support the associative rule for multiplication

Proof:

(1) By the definition of multiplication for real numbers, multiplication follows the properties of the set of rationals. [see Definition 7, here]

(2) So, the conclusion follows from the fact that the rational numbers support the associative rule for multiplication. [see Lemma 8, here]

QED

Lemma 7: The set of real numbers support the distributive rule

Proof:

(1) The properties of multiplication of reals is based on the properties of rational numbers [see Definition 7, here] and the properties of addition of reals is based on the properties of rational numbers [see Definition 5, here].

(2) So, the conclusion follows from the fact that the rational numbers support the associative rule for multiplication. [see Lemma 10, here]

QED

Lemma 8: The set of real numbers have an additive identity

Proof:

(1) The additive identity is the set of all rational numbers less than 0.

(2) Let x be any real number.

(3) The x+0 be the set of all rational numbers less than x added to all rational numbers less than 0.

(4) Let a be any rational number less than x.

(5) Let b be any rational number less than 0 so that be must be negative.

(6) a + b is thus less than a which is less than x.

(7) Since b can be as close to 0 as we want, a+b can be as close to x as we want.

QED

Lemma 9: The real numbers support an additive inverse property

Proof:

(1) Let x be any real number

(2) Let -x be the set of rational numbers that are less than -x.

(3) Using the definition for addition (see Definition 5, here):

x+-x is the set of all rational numbers less than 0.

QED

Lemma 10: The set of real numbers supports a multiplicative identity property

Proof:

(1) Let x be any real number

(2) The multiplicative inverse is 1 which is the set of rational numbers less than 1.

(3) It is clear that x*1 = {the set of rational numbers less than x } = x.

QED

Lemma 11: The set of real numbers supports a multiplicative inverse property

Proof:

(1) Let x = be any nonzero real number

(2) The multiplicative inverse is 1/x

(3) This is clear since the set defined by x*1/x is the set of all rational numbers less than 1.

QED

Theorem 12: The real numbers form a field

(1) The real numbers are closed on addition [see Lemma 1 above] and multiplication [see Lemma 2 above].

(2) The real numbers support the commutative property of addition [see Lemma 3 above], the associative property of addition [see Lemma 4 above], the commutative property of multiplication [see Lemma 5 above], an associative property of multiplication [see Lemma 6 above], and a distributive property [see Lemma 7 above].

(3) The set of real numbers has an additive identity property [see Lemma 8 above], an additive inverse property [see Lemma 9 above], a multiplicative identity property [see Lemma 10 above], and a multiplicative inverse property [see Lemma 11 above].

(4) From all these properties, the real numbers form a field. [see Definition 3, here]

QED

## Sunday, September 20, 2009

### The Set of Rational Numbers

In today's blog, I show how to formally construct the set of rational numbers from the set of integers.

I will then give a proof that the set of rational numbers forms a field. If you need a review of fields, check out here.

Definition 1: Set of rational numbers

We can define the set of rational numbers as the ordered pair of integers (a,b) where a,b are integers and b ≠ 0.

(a,b) + (c,d) = (ad + bc, bd)

Definition 3: Multiplication of rationals

(a,b) * (c,d) = (ac,bd)

Definition 4: Equality of rationals

Two rational numbers (a,b) and (c,d) are equal if and only if ad=bc.

Definition 5: Comparison of rationals

(a,b) is less than (c,d) if and only if abd2 is less than b2cd.

Lemma 1: All integers can be represented as rational numbers

(1) Any integer x can be represented as (x,1) [see definition 1 above]

(2) This correspondence holds over addition

x + y = z if and only if (x,1) + (y,1) = (x*1+y*1,1*1) = (x+y,1) = (z,1)

(3) This correspondence holds over multiplication

x*y = z if and only if (x,1)*(y,1) = (xy,1*1) = (xy,1) = (z,1)

QED

Lemma 2: The set of rational numbers is closed on addition

Proof:

(1) The integers are closed on addition [see Lemma 1, here] and multiplication [see Lemma 2, here].

(2) So, it follows from Definition 2 above that the rational numbers are closed on addition.

QED

Lemma 3: The set of rational numbers is closed on multiplication

Proof:

The follows directly from Definition 3 and the fact that the integers are closed on multiplication [see Lemma 2, here].

QED

Lemma 4: The set of rational numbers satisfy the Commutative Rule for Addition

Proof:

(1) From Definition 2 above:

(a,b) + (c,d) = (ad + bc, bd)

(2) Since integers are commutative by addition (see Lemma 7, here):

(3) From Definition 2 again, we get:

(bc + ad,bd) = (c,d) + (a,b)

QED

Lemma 5: The set of rational numbers satisfy the Associative Rule for Addition

Proof:

From Definition 2 above:

[(a,b) + (c,d)] + (e,f) = (ad+bc,bd) + (e,f) = (adf+bcf + bde,bdf) = (a,b) + (cf + de,df) =

= (a,b) + [(cf + de,df)] = (a,b) + [(c,d) + (e,f)]

QED

Lemma 6: The set of rational numbers has an Additive Identity for all elements.

Proof:

(1) Using Definition 2 above, we have:

(a,b) + (0,c) = (a*c + 0*b,b*c) = (ac,bc)

(2) Using Definition 4 above, we note that:

(ac,bc) = (a,b) since acb = bca [using the commutative property multiplication for integers, see Lemma 8, here]

QED

Lemma 7: The set of rational numbers has an Additive Inverse for all elements.

Proof:

(1) Let (a,b) be a rational number.

(2) Then, it's additive inverse is (-a,b) since:

(a,b) + (-a,b) = (a*b + -a*b,b*b) = (ab-ab,b*b) = (0,b*b)

QED

Lemma 8: The set of rational numbers supports the Associative Rule for Multiplication

Proof:

Using Definition 3 above, we have:

[(a,b)*(c,d)]*(e,f) = [(ac,bd)]*(e,f) = (ace,bdf) = (a,b)*[(ce,df)] = (a,b)*[(c,d)*(e,f)]

QED

Lemma 9: (c,c) = (1,1)

Proof:

The follows directly from definition 4 above since:

c*1 = 1*c

QED

Lemma 10: The set of rational numbers supports the Distributive Rule

Proof:

(1) Using Definition 2 and Definition 3 above, we have:

(a,b)[(c,d) + (e,f)] = (a,b)*(cf+de,df) = (acf + ade,bdf)

(2) Using Definition 3 above and Lemma 8 above, we have:

(acbf + aebd,bdbf) = (b,b)*(acf + aed,bdf) = (1,1)*(acf + aed,bdf) = (acf + aed,bdf)

(3) Using Definition 2 above, we have:

(ac,bd) + (ae,bf) = (acbf + aebd,bdbf)

QED

Lemma 11: The set of rational numbers supports the Commutative Rule for Multiplication

Proof:

(1) Using Definition 3 above, we have:

(a,b)*(c,d) = (ac,bd)

(2) Using the Commutative Property of Multiplication for Integers (see Lemma 8, here):

(ac,bd) = (ca,db)

(3) Using Definition 3 above, we have:

(ca,db) = (c,d)*(a,b)

QED

Lemma 12: The set of rational numbers has a Multiplicative Identity

Proof:

For any rational number (a,b), we have (see Definition 3 above):

(a,b)*(1,1) = (a*1,b*1) = (a,b)

QED

Lemma 13: For every nonzero element, the set of rational numbers has a Multiplicative Inverse

Proof:

(1) Let (a,b) be any rational number.

(2) Then (b,a) will be its multiplicative inverse since:

(a,b)*(b,a) = (ab,ba)

(3) Using the Commutative Property of Multiplication of Integers (see Lemma 8, here), we have:

(ab,ba) = (ab,ab)

(4) Using Lemma 8 above, we have (ab,ab)=1.

QED

Theorem 14: The set of rational numbers forms a field

Proof:

This follows directly from Lemma 2 through Lemma 13 above and from Definition 3, here.

QED

References