Placeholder Image

字幕列表 影片播放

  • Suppose you want to solve the equation X² - 4 = 0.

  • In regular algebra you would factor the left-hand side giving you (x-2)(x+2)=0.

  • Here the product of two terms is zero, so one of the terms must be zero.

  • If x–2=0, then X = 2.

  • And if x +2 = 0, then X = -2.

  • So this equation has two solutions: 2 and -2.

  • This seems straightforward, right?

  • But to solve this equation we used a BIG assumption:

  • that if the product of two terms is zero, then one of the terms must be zero.

  • In Abstract Algebra, this isn’t always true.

  • Let’s see an example where this technique of solving equations does NOT work.

  • Well solve the equation x² + 5x + 6 = 0 in the integers mod 12.

  • If you factor the left-hand side, you get (x+2)*(x+3) is congruent to 0 mod 12.

  • If you set each term to 0 mod 12, you get x=10 or x=9.

  • Are these all the solutions?

  • Since there are only 12 numbers in the ring of integers mod 12,

  • let’s go ahead and check all possibilities.

  • Well plug in 0 through 11 into the polynomial x² + 5x + 6 and see what we get.

  • If you plug in 0 you get 6, so x=0 is not a solution.

  • If you plug in 1 you get 12 which is congruent to 0 mod 12, so 1 IS a solution.

  • This is interesting.

  • Here’s a solution to the equation that we did NOT find by factoring.

  • Plugging in 2 gives us 8.

  • 3 gives you 6.

  • And let’s quickly fill out the table for the remaining numbers.

  • We see this equation has not two, but FOUR solutions: 1, 6, 9 and 10.

  • Factoring only identified the solutions 9 and 10.

  • So what went wrong?

  • To see the problem, let’s take a closer look at the equation.

  • After factoring, we have (x+2)*(x+3) is congruent to 0 mod 12.

  • Setting each term to zero gave us two of the solutions: 9 and 10.

  • But why did we miss the other two?

  • To see why, look what happens when you plug in 1.

  • This gives you 3 × 4, which is 0 mod 12.

  • And if you plug in 6, you get 8 × 9 which is also 0 mod 12.

  • We now see the culprit.

  • Working mod 12, it’s possible to multiply two non-zero numbers together and get 0.

  • We say the integers mod 12 havezero divisors,”

  • and it’s the zero divisors which make solving equations more difficult.

  • With real and complex numbers, this doesn’t happen.

  • These fields do not have zero divisors.

  • That’s why in regular algebra, you learn to solve many equations by factoring

  • and setting the terms to zero.

  • But in abstract algebra, this technique is not good enough.

  • I’d like to take a moment and talk about the termzero divisors.”

  • This name was chosen because division can be defined in terms of multiplication.

  • For example, when working with integers, we say B divides A

  • if B × C = A for some integer C. From this definition, the termzero divisorsmakes sense.

  • In the integers mod 12, 3 × 4 = 0, so both 3 and 4 divide 0.

  • They really arezero divisors.”

  • Let’s return to the equation X² + 5X + 6 = 0,

  • except this time well work mod 11 instead of mod 12.

  • The integers mod 11 do NOT have any zero divisors.

  • This is because 11 is a prime number.

  • The only way the product of two numbers is a multiple of 11

  • is if one of the numbers is divisible by 11.

  • And none of the integers 1 through 10 is a multiple of 11,

  • so this ring does not have any zero divisors.

  • Like before, we can factor the left-hand side giving us

  • (X + 2)*(X + 3) is congruent to 0 mod 11.

  • Setting each term to zero gives us x = 9 and x = 8.

  • Let’s see if there are any other solutions by plugging in all the integers mod 11.

  • When x = 0, x² + 5x + 6 is equal to 6.

  • Plugging in x = 1 gives us 1.

  • x = 2 gives us 9.

  • Let’s go ahead and fill out the rest of the table.

  • We see that the only solutions are 8 and 9, the two solutions we found by factoring.

  • This is a bit of good news.

  • When a ring R does NOT have any zero divisors, the traditional technique of solving an equation

  • by factoring and setting the terms to zero DOES work.

  • For this reason, there’s a term for such rings: integral domains.

  • Here’s the complete definition: an integral domain is a commutative ring R

  • with a multiplicative identity 1 and NO zero divisors.

  • A natural question is why do we require an integral domain to be commutative?

  • After all, it seems that the most important thing, at least for solving equations,

  • is there are no zero divisors.

  • One reason, is when the idea of an integral domain was developed, much of the focus was

  • on generalizing the integers for use in number theory.

  • Hence the wordintegralinintegral domain.”

  • There is, however, a term for an arbitrary ring with no zero divisors: a DOMAIN.

  • But youll encounter integral domains more frequently.

  • Integral domains have another useful property: the cancellation property.

  • Suppose you want to find the solutions to the equation 2X = 6Y.

  • Your instinct might be to first simplify this by dividing both sides by 2,

  • giving you x = 3y.

  • But when working with a ring, you may not be able to divide.

  • So a different approach would be to factor both sides, then cancel the 2s,

  • giving you x = 3y.

  • Factoring and cancelling does not require division.

  • But can you safely cancel in any ring?

  • Unfortunately, no.

  • Look at the equation 3x = 6y in the ring of integers mod 12.

  • This equation has multiple solutions, but one solution is x = 6 and y = 1.

  • This is because 18 is congruent to 6 mod 12.

  • But look what happens if you factor out 3 then cancel.

  • We get x = 2y.

  • Now if you plug in x=6 and y=1 you get 6 is congruent to 2 mod 12, which is false.

  • So in the original equation, 6 and 1 is a solution, but after cancelling, it is not.

  • So, when can you safely cancel??

  • To find out, suppose a*x = a*y in some ring R, and 'a' is not 0.

  • If we get everything on one side, we can factor out 'a' using the distributive property.

  • If the ring R does NOT have any zero divisors, then we know that x – y must be 0

  • because 'a' isn’t 0.

  • This implies that x = y.

  • In other words, we can cancel 'a' from both sides.

  • So another benefit of an integral domain is the cancellation property.

  • To recap, an integral domain is a commutative ring R with 1, that has no zero divisors.

  • The absence of zero divisors means you can use the cancellation property.

  • It also means you can solve an equation by factoring and setting the terms to zero.

  • Here’s a puzzle for you to discuss: how many different quadratic equations are there mod 12?

  • And do any of them have exactly two solutions?

  • And what’s the deal with integral domains? (laughter)

  • Theyre neither an integral, nor a function's domain? (laughter)

  • I mean, come on, you know?

  • And how about those viewers who haven’t subscribed?

  • I mean it’s free, it’s easy?

  • What’s stopping them?? (laughter)

Suppose you want to solve the equation X² - 4 = 0.

字幕與單字

影片操作 你可以在這邊進行「影片」的調整,以及「字幕」的顯示

B1 中級

積分域(抽象代數) (Integral Domains (Abstract Algebra))

  • 3 0
    林宜悉 發佈於 2021 年 01 月 14 日
影片單字