The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard.

next © 2007 Herbert I. Gross next In this Lesson we will deal with constraints that behave “badly”. That is, there are times when (1) what looks like a constraint isn’t really a constraint, and (2) a constraint leads to a contradiction. For example, suppose we want to find two numbers, x and y; and we know that x + y = 10. That is, x + y = 10 is one constraint. Since there are two variables we need one more constraint.

© 2007 Herbert I. Gross One choice for the second constraint might be x – y = 2. This leads to the system of equations next x + y = 10 x – y = 2 And by the method of the previous lesson, we could add the bottom equation to the top equation to conclude that 2x = 12, and therefore that x = 6. Knowing that x must equal 6, the top equation tells us that y must equal 4. Case 1

next © 2007 Herbert I. Gross However, now suppose that the second constraint had been 2x + 2y = 20. In this case the system of equations would be… next x + y = 10 2x + 2y = 20 Again, using the method of the previous lesson we could multiply both sides of the top equation by - 2 to obtain the equivalent system of equations… Case 2 - 2x + - 2y = - 20 2x + 2y = 20

next © 2007 Herbert I. Gross And if we now add the two equations… next {(x,y):0 = 0} = the set of all numbers Case 2 In the language of sets… - 2x + - 2y = - 20 2x + 2y = 20 0 + 0 = 0 we obtain the statement 0 = 0; which is true for all values of x and y. next

© 2007 Herbert I. Gross The reason for this is that the two equations in… …are equivalent; meaning that they are two different ways of expressing the same constraint. That is, if x + y = 10 then 2x + 2y = 2(x + y) = 2(10) = 20 Case 2 x + y = 10 2x + 2y = 20

next © 2007 Herbert I. Gross On the hand, suppose the second constraint had been 2x + 2y = 22. In this case, the system of equations would be… next And we could again multiply both sides of the top equation by - 2 to obtain the equivalent system of equations… Case 3 x + y = 10 2x + 2y = 22 - 2x + - 2y = - 20 2x + 2y = 22

next © 2007 Herbert I. Gross And if we now add the two equations… next {(x,y):0 = 2} = Case 3 (the empty set) 0 In the language of sets… - 2x + - 2y = - 20 2x + 2y = 22 0 + 0 = 2 we obtain the statement 0 = 2; which is false for all values of x and y. next

© 2007 Herbert I. Gross next The reason for this is that the two equations in… Case 3 x + y = 10 2x + 2y = 22 …are contradictory; meaning that by the rules of our “game” if x + y = 10, then 2x + 2y has to be 20. This is contrary to the fact that we are told that in this problem 2x + 2y has to be 22.

next © 2007 Herbert I. Gross The geometric representation of a 2-dimensional linear constraint is a straight line. Thus, two such constraints are represented by two straight lines. next Given any two straight lines there are only two possibilities; namely (1) the two lines have different slopes (that is, m 1 ≠ m 2 ) or (2) they have the same slope. Some Geometric Notes

next © 2007 Herbert I. Gross If they have different slopes, the two lines are not parallel, and hence, they intersect at one and only one point. This is the situation in Case 1 above. Namely, in the y = mx + b form, the line x + y = 10 in equivalent to y = - x + 10 ( = - 1x + 10); from which we see that its slope is - 1. next And the line x – y = 2 can be rewritten in the form y = x – 2; from which we see that its slope is 1. Our solution in Case 1 tells us that these two lines intersect at the point (6,4).

next © 2007 Herbert I. Gross An “easy” way to graph the line ax + by = c is by finding its x and y-intercepts and then drawing the straight line that is determined by the two intercepts. next Applying this to the equation x + y = 10, if we let x = 0, the equation becomes y = 10, and we see that the y-intercept is (0,10). In a similar way, setting y = 0 yields the equation x = 10; which shows us that (10,0) is the x-intercept. Note

next © 2007 Herbert I. Gross If the two lines have the same slope there are two possibilities. next One possibility is that the two lines are the same. This happened in Case 2 above, Namely x + y = 10 is one equation of the line that passes through the two points (10,0) and (0,10). And 2x + 2y = 20 is another equation of the line that passes through the same two points (10,0) and (0,10). Since no two different lines can pass through the same two points the two equations represent the same line.

next © 2007 Herbert I. Gross Pictorially… Case 2 (0,10) (10,0) next Since no two different lines can pass through the same two points the to equations represent the same line. y = - x + 10 2y = - 2y + 20

next © 2007 Herbert I. Gross The other possibility is that the two lines are parallel but different lines. In such a case the lines will contain no points in common. next This happened in Case 3 above. Namely, in the y = mx + b form, the line x + y = 10 in equivalent to y = - x + 10 ( = - 1x + 10); from which we see that its slope is - 1. And the line 2x + 2y = 22 can be rewritten in the form 2y = - 2x + 22 (or y = - 1x + 11); from which we see that its slope is also - 1.

© 2007 Herbert I. Gross This tells us that the two equations define two different lines that are parallel. next More specifically, the y = mx + b form tells us that one line passes through the point (0,10) and the other line passes through (0,11).

next © 2007 Herbert I. Gross Pictorially… Case 3 (0,10) (10,0) next (0,11) (11,0) y = - 1x + 11 This tells us that the two equations define different lines that are parallel. y = - x + 10 next Both slope are - 1.

next © 2007 Herbert I. Gross It is typical to use notation such as (6,4) as an abbreviation for the number pair, x = 6 and y = 4. In other words, each member of a 2-dimensional space can be represented as a pair of numbers (often referred to as a “2-tuple” by mathematicians). next On the other hand (6,4) is also used to denote a point in the xy-plane. The context tells us in which of the two ways the number pair is used. Note on Dimension

next © 2007 Herbert I. Gross However, the algebraic definition of dimension goes beyond its use in traditional geometry. next For example, suppose we have a simultaneous system of, say, 5 linear equations for which the solution is x 1 = 2, x 2 = 7, x 3 = - 4, x 4 = 3, and x 5 = 1. We may abbreviate this solution by the 5-tuple (2,7, - 4,3,1); and in this context it is a member of a 5-dimensional “space”. Note on Dimension

next © 2007 Herbert I. Gross Thus, even though it has no geometric equivalent, a 5-dimensional space is quite “normal” in algebra. next As an interesting aside, in science fiction stories we often hear time referred to as the fourth dimension. Mathematically, this makes very real sense. Note on Dimension

next © 2007 Herbert I. Gross For example, suppose we want to measure the temperature in a room. Clearly, the temperature will depend on where the thermometer is placed. Position in space is represented by the 3-tuple (x,y,z). Moreover, even at a fixed point in the room, the temperature may depend on the time (t) the temperature was recorded. Therefore, we if let T denote the temperature, we may write that T = f(x,y,z,t) and say that the domain of f is a 4-dimensional space. Note on Dimension

next © 2007 Herbert I. Gross While it’s nice to have a geometric way to visualize the solution to a system of linear equations, the fact is that for dimensions greater than 3 there is no visual way to do this. next Therefore, it is often necessary to describe the conditions algebraically for when there is a unique solution and when there isn’t.

© 2007 Herbert I. Gross That is: we shall try to solve the system… next a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 where a 1, a 2, b 1, b 2, c 1, and c 2 are constants. So let’s see how the algebraic version works in the case of a system of two linear constraints. To make our demonstration as rigorous as possible, we will replace the specific coefficients by more general ones.

next © 2007 Herbert I. Gross a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 To imitate what we did in our previous problems, let’s begin by multiplying the top equation of our system by a 2, a 1 a 2 x + b 1 a 2 y = c 1 a 2 and the bottom equation by – a 1 to obtain the system… - a 1 a 2 x + - a 1 b 2 y = - a 1 c 2 next

© 2007 Herbert I. Gross If we now add the bottom equation of our system to the top equation and simplify, we see that… a 1 a 2 x + b 1 a 2 y = c 1 a 2 - a 1 a 2 x + - a 1 b 1 y = - a 1 c 1 next (b 1 a 2 – a 1 b 2 ) y = c 1 a 2 – a 1 c 2 0 + b 1 a 2 y + - a 1 b 2 y = c 1 a 2 – a 1 c 2 b 1 a 2 y – a 1 b 2 y = c 1 a 2 – a 1 c 2 next

© 2007 Herbert I. Gross If b 1 a 2 – a 1 b 2 ≠ 0, we can divide both sides of (b 1 a 2 – a 1 b 2 ) y = c 1 a 2 – a 1 c 2 by it to determine that the value of y must be… next c 1 a 2 – a 1 c 2 (b 1 a 2 – a 1 b 2 ) …and we may use this value for y in the top equation a 1 a 2 x + b 1 a 2 y = c 1 a 2 to solve for the value of x.

next © 2007 Herbert I. Gross Because we aren't able to divide by 0, the “trouble” occurs when… next b 1 a 2 – a 1 b 2 = 0 The condition above is usually written in the equivalent form… a 1 b 2 – b 1 a 2 = 0

next © 2007 Herbert I. Gross To evaluate it, we multiply the term in the upper left by the term in the lower right to get the first term of the determinant (a 1 b 2 ), next In terms of a diagram… a1a1 a2a2 b1b1 b2b2 and then we subtract the product of the term in the upper right and the term in the lower left (a 2 b 1 ). a1b2a1b2 -a2b1-a2b1

© 2007 Herbert I. Gross The determinant above is called a 2 by 2 determinant. A determinant always has the same number of rows and columns because it tells us about the solution for a system of n linear equations containing n “unknowns”. Thus, if we had 4 linear equations in 4 unknowns the determinant of the system would be 4 by 4. a 1 b 1 a 2 b 2

next © 2007 Herbert I. Gross The word “determinant” was used to indicate that it determined what type of solution the system of equations had. next More specifically, if the determinant of the system is not 0, there is one and only one n-tuple that satisfied each of the equations (constraints). Note And if the determinant is 0, there are either no such n-tuples or else there are an infinite number of them.

© 2007 Herbert I. Gross We have already seen that if the two lines represented by the two equations… Geometric Interpretation of a 2 by 2 Determinant a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 …have equal slopes, the system of equations will not have a unique solution.

next © 2007 Herbert I. Gross There are several ways to compute the slope of a line; one of which is to rewrite the equation of the line in the y = mx + b form. In this way we see that… If a 1 x + b 1 y = c 1, then b 1 y = - a 1 x + c 1, or… + y = -a1b1-a1b1 c1b1c1b1 Hence, the slope of the first line is… m 1 = -a1b1-a1b1 next

© 2007 Herbert I. Gross Using the same approach we see that the slope m of the second line is… Hence, the two lines have the same slope if and only if… from which we see that… m 2 = -a2b2-a2b2 next -a2b2-a2b2 -a1b1-a1b1 = - a 1 b 1 = - a 2 b 2

next © 2007 Herbert I. Gross If we multiply both sides of the equation - a 1 b 1 = - a 2 b 2 by - 1, we obtain… or… …and by definition of a 2 by 2 determinant, the equation a 1 b 1 – a 2 b 2 = 0 is equivalent to… next a 1 b 1 = a 2 b 2 a 1 b 1 – a 2 b 2 = 0 a 1 b 1 a 2 b 2 = 0

next © 2007 Herbert I. Gross Computing the value of an n by n determinant when n is greater than 2 is beyond the scope of this course. However, the gist is that if we add two or more equations, the sum doesn’t give us any additional information. For example, suppose we start with the two constraints x + y + z = 10 and 2x + 3y + 4z = 26. If we add them, we get the equation 3x + 4y + 5z = 36. next

© 2007 Herbert I. Gross In other words if 3x + 4y + 5z = c and c ≠ 36, the solution set for the system of equations For example, there are no values for x, y and z for which it is true that x + y + z = 10, 2x + 3y + 4z = 26 and 3x + 4y + 5z = 40. next x + y + z = 10 2x + 3y + 4z = 26 3x + 4y + 5z = c will be the empty set.

next © 2007 Herbert I. Gross However, suppose we had just been presented with the system of equations without having been told how they were chosen. We could have proceeded in exactly the same way we did in Part 2 of Lesson 21. next x + y + z = 10 2x + 3y + 4z = 26 3x + 4y + 5z = 40 Namely, what we did was to develop a systematic way to replace the system by an equivalent but an easier to solve system.

next © 2007 Herbert I. Gross The least common multiple of the coefficients of x is 6. next x + y + z = 10 2x + 3y + 4z = 26 3x + 4y + 5z = 40 6x + 6y + 6z = 60 - 6x + - 9y + - 12z = - 78 - 6x + - 8y + - 10z = - 80 This results in the equivalent system of equations… Hence, we will multiply both sides of the first equation by 6; both sides of the second equation by - 3; and both sides of the third equation by - 2. next

© 2007 Herbert I. Gross next 6x + 6y + 6z = 60 - 6x + - 9y + - 12z = - 78 - 6x + - 8y + - 10z = - 80 In the above system, we replace the second equation by the second equation plus the first equation… next 6x + 6y + 6z = 60 - 6x + - 9y + - 12z = - 78 0 + - 3y + - 6z = - 18

next © 2007 Herbert I. Gross next 6x + 6y + 6z = 60 0 + - 3y + - 6z = - 18 - 6x + - 8y + - 10z = - 80 Then, we replace the third equation by the third equation plus the first equation. next 6x + 6y + 6z = 60 - 6x + - 8y + - 10z = - 80 0 + - 2y + - 4z = - 20

next © 2007 Herbert I. Gross next 6x + 6y + 6z = 60 0 + - 3y + - 6z = - 18 0 + - 2y + - 4z = - 20 This gives us the equivalent system… next We then simplify the system above by dividing both sides of the first equation by 6; x + y + z = 10 y + 2z = 6 y + 2z = 10 both sides of the second equation by - 3; and both sides of the third equation by - 2 to obtain… next

© 2007 Herbert I. Gross The last two equations in our system indicate the contradiction. Namely, according to the second equation y + 2z must equal 6; but according to the third equation it must also equal 10. This would imply that 6 = 10, which is obviously false. x + y + z = 10 y + 2z = 6 y + 2z = 10

next © 2007 Herbert I. Gross On the other hand, if we had used the same approach with the system… x + y + z = 10 y + 2z = 6 …the sequence of steps would have been the same except that system would have been replaced by… next x + y + z = 10 2x + 3y + 4z = 26 3x + 4y + 5z = 36

next © 2007 Herbert I. Gross next Since this time theyyy second and thirdyyyy equations are the same, we essentially have a system of 2 equations with 3 unknowns, namely… x + y + z = 10 y + 2z = 6 x + y + z = 10 y + 2z = 6 We have 1 degree of freedom in our system. For example, we can choose any value we wish for x and then solve the system for y and z.

next © 2007 Herbert I. Gross If we choose to let x = 0, our system becomes… …and if we now subtract the top equation from the bottom equation, next x + y + z = 10 y + 2z = 6 z = - 4; we see that z = - 4; and knowing that z = - 4 we see from either equation above that y = 14. 0 next Hence, one solution for… Is x = 0, y = 14 and z = - 4; x + y + z = 10 2x + 3y + 4z = 26 3x + 4y + 5z = 36 or in 3-tuple notation (0, 14, - 4). next

© 2007 Herbert I. Gross If you are intimidated by subtracting the top equation from the bottom equation, simply interchange the two equations. next For example… Aside y + z = 10 y + 2z = 6 …can be rewritten in the equivalent form… …whereupon we can subtract the bottom from the top to conclude that z = - 4. y + z = 10

next © 2007 Herbert I. Gross However, there are times when it is important not to interchange the order of the equations. In that case, we make use of the “add the opposite” rule. Namely, we may multiply both sides of the top equation by - 1, and then add the two equations. Aside

next © 2007 Herbert I. Gross next Aside For example… y + z = 10 y + 2z = 6 …can be rewritten in the equivalent form… - y + - z = - 10 y + 2z = 6 …whereupon we can add the two equations to conclude that z = - 4.

next © 2007 Herbert I. Gross If we choose to let x = 1, our system becomes… …and if we now subtract the top equation from the bottom equation, next x + y + z = 10 y + 2z = 6 z = - 3; we see that z = - 3; and knowing that z = - 3 we see from either equation above that y = 12. 1 next Hence, another solution for… Is x = 1, y = 12 and z = - 3; x + y + z = 10 2x + 3y + 4z = 26 3x + 4y + 5z = 36 or in 3-tuple notation (1, 12, - 3). 9 next Returning now to our given problem…

© 2007 Herbert I. Gross More generally, we may choose any value for x that we wish. For example… next If we choose to let x = c, our system becomes… x + y + z = 10 y + 2z = 6 z = - 4 + c – c …and if we now subtract the top equation from the bottom equation, we see that z = - 4 + c next

© 2007 Herbert I. Gross next We may then find the value of y by replacing z by - 4 + c in this equation y + 2z = x to obtain… y + z = 10 – c y + 2z = 6 z = - 4 + c y + 2 ( - 4 + c) = 6 y + - 8 + 2c = 6 y + 2c = 14 y = 14 – 2c z = c – 4 In 3-tuple format, for each choice of c, the 3-tuple (c, 14 – 2c, c – 4 is a member of the solution set). next

© 2007 Herbert I. Gross As a check, we replace x = c, y = 14 – 2c, and z = c – 4 and obtain… next c + y + z = c + (14 – 2c) + (c – 4) = (c + c – 2c) + (14 – 4) = 10 And since there are infinitely many choices for c, the system has infinitely many solutions, one for each choice of c. As a quick check we see that if c = 0 the 3-tuple is (0, 14, - 4), and if c = 1 the 3-tuple is (1, 12, - 3). This agrees with the results we obtained previously. next 2c + 3y + 4z = 2c + 3(14 – 2c) + 4(c – 4) = 2c + 42 – 6c + 4c – 16 = 26 3c + 4y + 5z = 3c + 4(14 – 2c) + 5(c – 4) = 3c + 56 – 8x + 6c – 20) = 36 next x + y + z = 10 2x + 3y + 4z = 26 3x + 4y + 5z = 36

next © 2007 Herbert I. Gross In summary, the fact that we have 1 degree of freedom means that given the system of equations… …we may omit any one of the three given equations without changing the solution set of the system. next x + y + z = 10 2x + 3y + 4z = 26 3x + 4y + 5z = 36

The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard.

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The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard.

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Presentation on theme: "The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard."— Presentation transcript:

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