Interesting Questions, Facts and Information
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Interesting Questions, Facts, and Information
C2 x C2. The Galois group is the group of automorphisms of a field within an extention field. For example, take the real numbers (R) and the extension field of the complex numbers (C). Map each complex number to its conjugate (i.e. change the sign of the imaginary part) so for example 2+8i becomes 2-8i. Notice that the real numbers don't change since there is no imaginary part (-5.6 maps to -5.6) so this mapping would be an automorphism of the reals in the complex field. It is to Evariste Galois credit that he recognized that the structure of this group can answer many (previously) unsolved problems in mathematics.
The trick is to show that Q[sqrt(2)+ sqrt(3)] = Q[sqrt(2), sqrt(3)]. Clearly Q[sqrt(2), sqrt(3)] has sqrt(2)+sqrt(3) as an element, so Q[sqrt(2)+ sqrt(3)] is a subfield of Q[sqrt(2), sqrt(3)]. To go the other way, [sqrt(2)+ sqrt(3)]^3 = 11[sqrt(2)] + 9[sqrt(3)]. Subtract 9[sqrt(2)+ sqrt(3)] so 2[sqrt(2)] (hence sqrt(2)) is an element of Q[sqrt(2)+ sqrt(3)] and therefore so is sqrt(3). Thus Q[sqrt(2), sqrt(3)] is a subfield of Q[sqrt(2) + sqrt(3)]. By double inclusion, the fields are the same.
Whew! Now the easy part. By other theorems in Galois theory we can establish that in this problem, sqrt(2)is mapped to +/- sqrt(2) and sqrt(3) is mapped to +/- sqrt(3). This means the Galois group is C2xC2.
6. N.B.: sqrt stands for "square root of". cbrt stands for "cube root of".
The splitting field is an extension field that contains all of the roots the polynomial. For example, The polynomial 5X+1 has a rational number as its root, so the splitting field is the field of rational numbers (abbreviated as Q). For X^2-2, the roots are +/- sqrt(2) so we need to take the rationals and generate an extension field by that contains sqrt(2). This means the splitting field is Q[sqrt(2)].
Degree is just that - the degree of the polynomial that has an element of the field as a root. For example, what is the degree of Q(cbrt(3)) over Q? The polynomial that has the smallest degree with rational coefficients with cbrt(2) as a root is X^3-2. This polynomial had degree 3 so the degree of Q(cbrt(3)) over Q (written as [Q(cbrt(3)):Q] is 3.
The splitting field contains all three roots of X^3 - 2, specifically cbrt(2), w[cbrt(2)], and w^2[cbrt(2)] where w is [-sqrt(3)]/2 + (1/2)i. We note that Q(cbrt(2), w) as an extension field contains all 3 roots. As stated above, E=Q[cbrt(2)] is degree 3 over Q but cbrt(2) is real and cannot account for the needed complex numbers. The smallest polynomial with elements of Q[cbrt(2)] for coefficients AND w as a root is X^2 + X + 1 so F=E(W) is degree 2 over E. Putting everything together, [Q(cbrt(2)),w:Q] = [Q(cbrt(2)),w:Q(cbrt(2))] x [Q(cbrt(2)):Q] = 2 x 3 = 6.
An integral domain. An example of a ring that is not an integral domain are the integers modulo 6 (the remainder when a number is divided by 6) abbreviated as Z. This is not an integral domain because 2x3=6 which is 0 (mod 6) but neither 2 nor 3 are 0 (mod 6). The ring Z[p] with p being a prime number is always an integral domain because the only factors of a prime number are 1 and itself.
Based on this logic, it should be pretty easy for you to show that Z[c] with c being a composite number is NEVER an integral domain. Give it a try. Send me a message if you need a hint.
The complex numbers. An extension field is made by taking a field adding one or more elements and creating a new field using the properties discussed in Question One. The complex numbers start out with the field of real numbers and the square root of -1 (i). Since the field has to be closed under multiplication, we have to include 2i, 3i, (4/91)i, etc. in the field. The inverses have to be included as well, both additive (-i, -2i, -3i . . .) and mutiplicative (i/2, i/7, i/(-193/16), etc.) and closed under addition (1+9i, -4+6i, 2-17i). This procedure is repeated with all of the new elements. Eventually, this leads to a definition of complex numbers: the set of all numbers in the form A+Bi where A and B are real numbers. Thus the complex numbers are an extension field of the real numbers.
This question/answer is a fancy way to say that the every root of every polynomial with real coefficients is a complex number. The Fundamental Theorem of Algebra guarantees that these roots do exist. More over, the number of roots equals the degree of the polynomial (i.e. the largest power of X). Interestingly enough, the complex numbers are an algebraically closed field which means a polynomial with complex coefficients has all of their roots in the complex field. This is why we don't go beyond complex numbers in high school. They form a kind of cul-de-sac in exploring polynomials.
S3. S3 is the symmetric group (every combination) of 3 elements. D6 is the dihedral group (rotating and flipping) of a hexagon. C3 x C3 are the ordered pair combining 0's, 1's, and 2's. A6 is the alternate 6 group which is formed by starting with (123456) and creating elements of S6 by applying an even number of transpositions (switching two adjacent numbers).
Symmetric groups are very important since an important theorem in algebra states that EVERY group is a subgroup of some symmetric group. They're also important in establishing an important result from Galois theory, but I'll save that for question 10.
Commutative multiplication. Rings include all of the common properties for addition (closure, commutative, associative, identity, and inverses) along with the distributive law, closure under multiplication, and associative multiplication. Closure means that you cannot move out of the set when performing the operation. For example, integers are not closed under division because 3 divided by 19 is NOT an integer. Special rings can include combinations of commutative multiplicative (commutative rings), identity (rings with identity), and inverse (division rings). If a ring has ALL of these properties it is called a field.
These rules form a natural progression when learning arithmetic. We learn about adding and multiplying natural numbers but natural numbers are very limited (you can't have a problem like 3-8). We then proceed to integers which is the ring formed by the natural numbers, zero, and their inverses. This new system lets us do more but we still have a problem with division. What is -17 divided by 2 using long division? Is it -9r+1 or -8r-1? We need to have multiplicative inverses and so we take the integers and create the field of rational numbers commonly known as fractions. Now you know why math after 5th grade is so dependent on fractions.
|Mr. Po jumps for joy when he gets TWO bottles of soda in a broken vending machine when pressed in one key, D1. Does this represent a function?||Algebra Wonderland!
No. A function occurs when there is only ONE y value, in this case is the soda bottles, for each x vale, in this case is the key D1. In this scenario, Mr. Po received two bottles, which stops this from being a function.
|Notve Rysmart is in debt of 6 dollars. In one year, his debt multiplies by the same amount. In the year after that, it again multiplies by the same amount he had to begin with. How much debt is Notve in now? ||Algebra Wonderland!
$216. $6 times $6 times $6 equals -$216. This could be represented by $6^3.
|Skinny Knee is drawing a graph representing the growth of his knees over the past year. He found that following equation, 4x+2y=6, could be used to represent the line in his graph. Using this information, which equation is perpendicular to his line?||Algebra Wonderland!
3x-6y=5. First convert the equation in the problem, 4x+2y=6, into y-intercept form. y=-2x+3 is the result. Perpendicular equations will have the opposite reciprocal of the slope. 2's opposite reciprocal is -1/2, so 3x-6y=5 is correct.
|The sum of two numbers is 74. The difference between the two numbers is 16. What are the two numbers?||Algebra Wonderland!
29 and 45 . Solve by using system of equations. The two equations are x+y=74 and x-y=16. Isolate either y or x in the second equation and substitute in the first.
|What is the y-intercept of a direct variation? A direct variation always passes through the origin.||Algebra Wonderland!
0 . A direct variation is a line that always passes through the origin (0,0)! The y-intercept is the point which the line crosses the y-axis.
|Let's start out with food! There is one chocolate mint bar with plenty of sprinkles on it. Obe Seguy and his 4 friends want to buy it. They buy it using their combined money that totaled to $2.50. If they decide to split it into equal parts for each one of them, what is the percentage of the whole chocolate bar each of the kids will receive?||Algebra Wonderland!
20% . There are five people (Obe with his 4 friends). Divide one (the whole chocolate bar) by five, and .2 is the answer. After converting this into a percent, the answer is 20%.
Both binomials and trinomials. Distracting as the prefix 'poly-' meaning 'many', 'more than one', or 'infinite number', polynomial refers to all expressions, including monomials (1 term), binomials (2 terms), and trinomials (3 terms), as well as expressions with more than 3 terms.
|You deposit $1000 in the bank at 4% compounded daily for 5 years. Mathematically, what do you call the '4%'? ||Algebra Terms
Growth rate. The initial amount is repeatedly multiplied (5 times) by the growth factor of 1.04, giving the final amount. The growth factor is the sum of '1' and the growth rate (.04 or 4%). Interest is the difference between the initial amount and the final amount. Profit is an accounting term.