Proof That Isometries Preserve Parallel Lines

AI Thread Summary
The discussion centers on proving that isometries preserve parallel lines. The initial proof attempted to show that if L and K are parallel lines, then their images F(L) and F(K) under an isometry F are also parallel. Critics pointed out that the proof lacked formality and did not adequately generalize the argument for all points on the lines. A suggested improvement involved using an indirect proof to demonstrate that if F(L) and F(K) were not parallel, it would lead to a contradiction regarding the distance between points on the lines. Ultimately, the revised proof clarified that F(L) and F(K) must indeed be parallel, reinforcing the preservation of parallelism by isometries.
bistan
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Hey guys, just wanted to get a quick check that this proof is sound.

Homework Statement


Let L, K be two parallel lines, and let F be an isometry. Prove that F(L) & F(K) are parallel.

Homework Equations


If P and Q are points on the plane and F is an isometry, the distance PQ = the distance F(PQ).

The Attempt at a Solution



Let P be a point on L and Q a point on K. By definition of parallel lines, L & K have no point in common. Because F is an isometry, the distance PQ = the distance F(PQ). Therefore F(L) & F(K) must also have no point in common. Thus F(L) & F(K) are parallel.
 
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Looks fine to me. Maybe should mention use of the fact that the distance between two points is zero if and only if they are the same point.
 
haruspex said:
Looks fine to me. Maybe should mention use of the fact that the distance between two points is zero if and only if they are the same point.

Maybe I'm missing something but I don't see anything in what bistan did that would be anything like a proof.
 
Dick said:
Maybe I'm missing something but I don't see anything in what bistan did that would be anything like a proof.
Maybe I'm reading too much between the lines :blushing:
It appears to lean on some standard facts like F(K), F(L) will be straight lines; parallel lines don't intersect; non-parallel lines do intersect. And I assume this is Euclidean (otherwise I'm not sure how parallel lines are defined).
 
Dick said:
Maybe I'm missing something but I don't see anything in what bistan did that would be anything like a proof.

It's not very formal I understand. I'm working through Serge Lang's Basic Mathematics and I'm using the facts that I've been given. Maybe I should've mentioned that.
 
I agree with Dick. Saying "Let P be a point on L and Q a point on K. By definition of parallel lines" means that P and Q are specific points on L and K. The fact that the distance between two specific points is the same as the distance between their images says nothing about the lines. You are trying to interpret that to say that the distance between any two points F(L) and F(K) is non-zero but that is not what you is means.

Better would be an indirect proof: if F(L) and F(K) are NOT parallel then there exist F(P) on F(L) and F(Q) on F(K) such that F(P)= F(Q)- that is they are the same point. That means the distance between them is 0. Now use the isometry to go back to L and K.
 
HallsofIvy said:
I agree with Dick. Saying "Let P be a point on L and Q a point on K. By definition of parallel lines" means that P and Q are specific points on L and K. The fact that the distance between two specific points is the same as the distance between their images says nothing about the lines. You are trying to interpret that to say that the distance between any two points F(L) and F(K) is non-zero but that is not what you is means.

Better would be an indirect proof: if F(L) and F(K) are NOT parallel then there exist F(P) on F(L) and F(Q) on F(K) such that F(P)= F(Q)- that is they are the same point. That means the distance between them is 0. Now use the isometry to go back to L and K.

Good thing I posted on here! That's more or less what I thought; that stating two arbitrary specific points would allow it to be made general for the whole line. Thanks for your help everyone.

So:

If F(L) and F(K) are not parallel then there exists point F(P) on line F(L) and point F(Q) on line F(K) such that F(P)= F(Q). Since F is an isometry and preserves distance, P = Q which is impossible since lines L and K are parallel. Thus F(L) and F(K) have no point in common and are parallel.
 
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bistan said:
Good thing I posted on here! That's more or less what I thought; that stating two arbitrary specific points would allow it to be made general for the whole line. Thanks for your help everyone.

So:

If F(L) and F(K) are not parallel then there exists point F(P) on line F(L) and point F(Q) on line F(K) such that F(P)= F(Q). Since F is an isometry and preserves distance, P = Q which is impossible since lines L and K are parallel. Thus F(L) and F(K) have no point in common and are parallel.

That looks better.
 
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