- #1
gumy77s
- 1
- 0
Consider the system [S, L, P], where S contains exactly four points A, B, C, and D, the lines are the sets with exactly two points, and the planes are sets with exactly three points. This "space" is illustrated by the following figure:
Here it should be remembered that A, B, C, and D are the only points that count. Show that under the incidence postulates, S cannot be a line.
Incidence Postulates:
I-0)All lines and planes are sets of points.
I-1) Given any two different points, there is exactly one line containing them.
I-2) Given any three different noncollinear points, there is exactly one plane containing them.
I-3) If two points lie in a plane, then the line containing them lies in the plane.
I-4) If two planes interesect, then their intersection is a line.
I-5) Every line contains at least two points. S contains at least three noncollinear points. Every plane contains at least three noncollinear points. And S contains at least four noncoplanar points.
-----
This is what I have thought about doing so far but I'm not sure if it is right or not. Is there anything I'm missing? Also, I'm completely confused about how to show that S cannot be a line using I-0. A push in the right direction would be greatly appreciated! Thank you in advance!:
I-1 states that given any two different points, there is exactly one line containing them. Since we have more than one set of two points in S, S consists of multiple lines and therefore cannot be a line.
I-2 states that given any three different noncollinear points, there is exactly one plane containing them. Noncollinear implies that the points do not all lie on the same line, as is the case with any group of three points in the figure. Any group of three points form a triangle, which is a set of three lines that join at endpoints to enclose a space.
I-3 states that if two points lie in a plane, the line containing them lies in a plane. We already know from I-2 that planes exist in the figure because we have four sets of three noncollinear points (ABC, ABD, ACD, and BCD). We know that A and B lie in a plane. Therefore, line AB lies in the plane. We can say the same for all other pairs of points in the system, and this shows that we have more than one line.
I-4 states that if two planes intersect, then their intersection is a line. We obviously have four planes in our system, as stated earlier. The intersection of any two planes consists of the lines connecting two points the planes have in common. For instance, ABC intersect ABD is line AB and ABC intersect ACD is line AC. Obviously, this is more than one line, so S cannot be a line.
I-5 states that tells us that every line consists of at least two points. This is shown in the figure, as we have AB, AC, AD, BC, BD, and CD - this is six lines, so again, S can't be a line.
Here it should be remembered that A, B, C, and D are the only points that count. Show that under the incidence postulates, S cannot be a line.
Incidence Postulates:
I-0)All lines and planes are sets of points.
I-1) Given any two different points, there is exactly one line containing them.
I-2) Given any three different noncollinear points, there is exactly one plane containing them.
I-3) If two points lie in a plane, then the line containing them lies in the plane.
I-4) If two planes interesect, then their intersection is a line.
I-5) Every line contains at least two points. S contains at least three noncollinear points. Every plane contains at least three noncollinear points. And S contains at least four noncoplanar points.
-----
This is what I have thought about doing so far but I'm not sure if it is right or not. Is there anything I'm missing? Also, I'm completely confused about how to show that S cannot be a line using I-0. A push in the right direction would be greatly appreciated! Thank you in advance!:
I-1 states that given any two different points, there is exactly one line containing them. Since we have more than one set of two points in S, S consists of multiple lines and therefore cannot be a line.
I-2 states that given any three different noncollinear points, there is exactly one plane containing them. Noncollinear implies that the points do not all lie on the same line, as is the case with any group of three points in the figure. Any group of three points form a triangle, which is a set of three lines that join at endpoints to enclose a space.
I-3 states that if two points lie in a plane, the line containing them lies in a plane. We already know from I-2 that planes exist in the figure because we have four sets of three noncollinear points (ABC, ABD, ACD, and BCD). We know that A and B lie in a plane. Therefore, line AB lies in the plane. We can say the same for all other pairs of points in the system, and this shows that we have more than one line.
I-4 states that if two planes intersect, then their intersection is a line. We obviously have four planes in our system, as stated earlier. The intersection of any two planes consists of the lines connecting two points the planes have in common. For instance, ABC intersect ABD is line AB and ABC intersect ACD is line AC. Obviously, this is more than one line, so S cannot be a line.
I-5 states that tells us that every line consists of at least two points. This is shown in the figure, as we have AB, AC, AD, BC, BD, and CD - this is six lines, so again, S can't be a line.