00:01
Let's define see the set of order pairs of real numbers, that is the x, y, pairs, where x and y are real numbers.
00:14
There, we define addition as is indicated here, the sum of two pairs.
00:28
Here is a pair, not a sum here.
00:34
Sum of two pairs is done term by term.
00:38
That is the resulting pair, the first component is the sum of the first components of the two pairs, and the second component of the resulting pair is the sum of the second components of the two pairs.
00:53
The multiplication is defined differently.
00:57
The first components of the product is the first two components, multiplication of two first components minus a multiplication of two and the second components.
01:10
That's the first component of the result, and the second component of the result is the first component of the first factor times the second component of the second factor, plus second component of the first factor times first component of the second factor.
01:32
And we recognize this multiplication here as defined for the complex number, where the second component is the imaginary part and the first component is the real part.
01:48
Then we will show that with respect to these operations and algebraic properties on page 3 are satisfied with are replaced by the set c.
01:58
So there are five properties.
02:02
We're going to name it, then name it as in the page 3 of the book.
02:07
So a, a1 is the associative property.
02:12
In this case, at least two properties are a direct consequence of the properties or the equivalent properties under real numbers.
02:27
So let's say a1 is associative properties of m.
02:30
X1 plus x2, y2, plus x2, y2, plus x3, y3, that is, we sum first the last two pairs, and the result is lastly add to the first pair.
03:00
We will show that this is equal to doing the first sum of the two pairs, and the result adding that to the third pair.
03:16
That is what we see here.
03:23
Y2 here so let's do this side of the equality and we will get to this so we have x1 y1 plus x2 y2 plus x3 is equal to x1 1 .1 plus and we add these two pairs we know is equal to x2 plus x2 plus x3 as the first component and we add these two pairs we know is equal to x2 plus x3 as the first component and we y2 plus y3 as a second component and then we have here the sum of two pairs and that's equal to the first component of the first pair x1 plus the two the first component is the second pair and the second component of the result is the second component of the first pair plus the second component of the second pair which is a parenthise here component that's it now we know the real numbers have verified the associative property for the sum and so this is equal to x1 plus x2 and that plus x3 and then this is equal to x1 plus x2 and then that plus y3 and then we recognize here we have x1 plus x2 y1 plus y2 plus y2 plus y2 plus this square bracket is not necessary here plus x3 y3 it is clear if we do this we get this and then this is equal to this result here sorry this result here this one is exactly the sum of x1 y1 plus x2 to y2.
06:06
And so we have proved that.
06:09
Doing the sum of the last two pairs and the result of that plus the first pair is the same as adding the first two pairs and the result of that plus the third pair.
06:23
So it's the associative property and it was a consequence directly of the same property on the real numbers.
06:30
So let's see a2, a2 is, okay, there's something i forgot here.
06:39
I verified the associative property for the product.
06:46
For the sum, for the product we have, okay, let's say a1, and here i should have put addition, and now heal multiplication, multiplication, and now we do the same, that is, but here we got a little bit, careful.
07:16
So i'm going to do first this x1 y1 times x2 y2 and that times x3 y3.
07:39
So i do first these two pairs and we know that x1 times x2 minus y1 y2 first components a component x1 plus y1 x2 and that times x3 y3 and here we got to be careful a bit so is first component here x1 x2 minus y1 y2 times x3 minus x1 y2 plus y1 x2 times y3 that's the first component of these product here okay then x1 x1 x2 minus y1 y2 times y3 plus x1 y2 x3 so that's the result of this way to three pairs.
09:14
First is two first pair pairs and then that times the third pair.
09:21
Well if we want, this is x1, x2, x3 minus y1, wight 2 x3 minus x1, y2, y2, y3 minus x2, y3, then this is x1, x1, x2, x2y3 minus y2 we 1 .2 .3 plus x1.
09:56
X3 y2 plus x2.
10:03
X2 x3y1.
10:05
Okay, let's verify a little bit.
10:17
Okay, okay, that's correct.
10:30
And now we do the other way of multiplication is x1 y1 times x2 y2 times x3 y3 so we do the first the last two pairs so this is x1 y1 times so we have x2 x3 minus y2 y3 and then x2 y3 plus y2 x3 and then x2 y3 plus y2 x3 and the doing this last one we get x1 x2 x3 minus y2 y3 minus y1 x2 y3 minus y1 x2 y3 plus y2 x3 first component and now second component is x1 x2 y2 plus y2 x3 that plus uh plus y1 times x2 x3 minus y2 y2 y 3 okay so that's x1 x2 x2 x3 minus x2 y2 y3 minus x2 y3 minus x2 y3 minus x2 y1 y3 minus x2 y3 minus x2 y3 minus x3 y1, y2, first component, second component, x1, x2, y3, plus x1, x3, y2, plus x2, y2, minus, y2, and y2, and y3, and y3.
12:53
And now let's see, okay, let's see if we have first component first.
13:07
So we have this, is this here, this one here, minus x1 way 2 or 3, minus x1 where 2 or 3 is here, minus x2 way 1 with 3, minus x2 1 wave 1 with 3, minus x2 1 wave 3, minus x2 1 wave 3, 3, minus x2 1 wave 3, 3, 3 minus x2, y1, y2 minus x3, y1, w2.
13:39
So the first component is exactly the same.
13:43
Let's see now the second component, x1, x2, y3 is here.
13:52
X1, x3, y2 is here.
13:59
X1, x3, y2, correct.
14:02
Now, negative, this term, x2, x3, y1, positive, x2, x3, y2, is this one here and it's only one left negative y1, 1 ,1, 2 ,0, is correct.
14:18
This one here is this one here.
14:21
So they are exactly equal.
14:27
So, x1, y1 times x2, y2 times x3, y3, is exactly the same as x1 y1 times x2 y2, that times x3 y3 so it is only a 1 so we move to a 2 so a 2 is a 2 is commutative property so let's see if we do 1 times x2 y2 we know it sorry the sum first that is x1 plus x2 y1 plus y2 and because the real numbers in the real numbers the sum or addition is commutative this is the same as x2 plus x1 and the second components is the same as y1 or y2 plus y1 and we recognize this is x2 y2 plus x1 so is directly true because the same property in the real numbers for the multiplication, there is more work to do as we saw above.
16:07
So we are going to do the product in this direction first, in this order.
16:12
So we get x1, x2 minus y1, y2, and then x1, y2 plus x, y1, x2.
16:31
And now we're doing the other order x2 y2 times x1, y2.
16:37
And we get x2 x1 minus y2 y1 and then x2 y1 and then x2 y1 plus y2 x1 and now we compare the two this one this one because the multiplication is commutative in the real numbers and this one is this one for the same reason now this one here is this one here for the same reason now this one here for the same reason of community for real numbers and this one here is this one here so it seems that we have proved directly here very quickly that the community property for the sum and multiplication holds and now we go for a 3 and a 3 is that there is zero for the sum any pair plus the pair which is represented presenting the 0 is equal to the pair, the given pair, and there is one which has a property that any pair times that pair representing one is the given pair.
18:09
So let's see, it seems that it's natural to define the 0 for this set c as the pair of two components equal to 0.
18:21
So let's see that that.
18:23
That is if we sum x1y1 any pair plus 0 0 you know this is x1 plus 0 y1 plus 0 but 0 has a property in the real numbers that any number any real number plus 0 is that real number so we get x1 and y1 again so the 0 or the additive 0 in c is is zero, zero.
19:08
The pair of the two components equal zero.
19:13
And we have proved that here directly.
19:16
And now for the product we see in this case is not the pair of both components equal to one, but one zero.
19:30
And that's because the imaginary part, which is the second part, it should be zero because we should have a one which is the same number in the real set, in the set of real numbers, because the imaginary part is zero, so we identified this pair with the real number one.
19:55
And that should be the one for the multiplication.
20:01
Here is a multiplication.
20:04
So minus two.
20:07
So we get, if we apply the way of multiplicates in c, we get x1 times.
20:13
Times 1 minus y1 times 0 and then x1 type 0 plus y1 times 1 but we get here this is 0 so we get x1 times 1 is x1 because 1 is the one element for the multiplication in the real numbers and then we have this is 0 and we get way 1 times 1 is way 1 so that's it so the multiplicative 1 in c is the pair 1 0 okay so a 4 now given any pair in c that should be another pair add to c to the original the given pair is 0 in this case you're 0 so it's clear that for for any x 1, y1 in c, we have that x1 y1 plus negative x1 negative y1 is 0 because x1 plus negative x1 is 0 because x1 plus negative x1 is 0 and y1 plus negative y1 is 0.
22:07
And this is the additive 0 in c.
22:12
So any pair in c has another pair.
22:17
Which add with the given pair, we obtain the zero, the additive zero in c.
22:25
Now for the multiplication, it's a little bit more complicated in the sense that given any pair, we got to find another pair that multiply with it.
22:38
We get the multiplicative one that is one zero in this case.
22:44
The idea behind this is to state the following.
22:49
We have x1y1 and we want to find x2 y2, that is this is given and we want to find this, such that the product is 1 -0.
23:03
That's the idea.
23:04
We are looking for the multiplicative inverse of any pair in the sense of the multiplication defined in c.
23:13
And for that, the multiplication of the two pairs to give us the multiplicative 1 in c, which is 1 -0...