Using the definition of isometries

Question

Suppose fRR is an isometry of the reals. Prove that f is a reflection in a point if and only if f is decreasing.

 

 

We are learning about isometries in class and I am struggling using them to help answer this question. We are supposed to use definition of isometries but I am having a hard time understanding the concept. If someone could help that would be great!

Answer 1

Megan,

 

Sorry, I didn't see that you'd posted your comment as a question as well! Here's the answer I gave on your other question:

 

 

 

Again, f is an isometry from R to R (of the real numbers). This time, though, we'd like to prove that f is a reflection in a point if and only if f is decreasing.

And again, since it's if and only if, you'll essentially do two proofs.

Useful math definitions:

Isometry <--> d(x, y) = d(f(x), f(y))
Decreasing <--> x < y implies f(x) > f(y)
Reflection over b <--> (x + f(x)) / 2 = b

That last one essentially says that the point b is the average of x and f(x). In other words, if you reflect a number over the point b, b will be exactly in between the number and its reflection.

(Try it out if that doesn't quite seem clear... Pick a number to reflect over, like 4, and a number to reflect, like -3. -3 will become 11, and indeed (-3+11)/2 = 8/2 = 4.)



1. PROVE THAT IF F IS A REFLECTION, THEN F IS DECREASING

[For this one, our toolbox is d(x, y) = d(f(x), f(y)) since f is an isometry, and (x + f(x)) / 2 = b and (y + f(y)) / 2 = b for some real number b since f is a reflection. We'll probably use both of these facts. And, notice that I just assign variables when I need to... x and y to stand for two random real numbers to put through our function and b to stand for the point f is reflecting real numbers over.

And remember, we'd like to prove that, if x < y, then f(x) > f(y). As soon as we do that, we're done with this part.]

Assume f is a reflection across the point b. Let x and y be two real numbers such that x < y. Then, since f is a reflection, (x + f(x)) / 2 = b and (y + f(y)) / 2 = b. Rearranging a bit, we get:

x + f(x) = 2b
f(x) = -x + 2b

y + f(y) = 2b
f(y) = -y + 2b

We know that x < y. This means that -x > -y.

[You can prove this if you feel like you need to. Think through several cases: x and y could both be negative or could both be positive, x could be negative and y could be 0, x could be negative and y could be positive, or x could be 0 or y could be positive. Reassure yourself that, in all these cases, -x > -y as long as x < y.]

By the addition property of inequality, since -x > -y, -x + 2b > -y + 2b. Thus f(x) > f(y). So, by definition, f is decreasing.



2. PROVE THAT IF F IS DECREASING, THEN F IS A REFLECTION

[For this one, our toolbox is d(x, y) = d(f(x), f(y)) since f is an isometry, and x < y implies f(x) > f(y) since f is decreasing.

We'd like to prove that (x + f(x)) / 2 = b and (y + f(y)) / 2 = b for some real number b, since that would mean f is a reflection over a point b. As soon as we do that, we're done with this part.]

Assume f is decreasing. Let x and y be two real numbers such that x < y. Then f(x) > f(y). Since isometries preserve distance by definition, this means that x - y = f(y) - f(x).

[Notice that this is almost the same phrase we used with the translation, but with f(y) and f(x) switched. That's because, this time, x and f(y) are the two smaller numbers, not x and f(x).]

Let b be the average of x and f(x). That is, b = (x + f(x)) / 2. Then 2b = x + f(x), and 2b - f(x) = x. Substituting into the above equation, we get:

2b - f(x) - y = f(y) - f(x)
2b - y = f(y)
2b = y + f(y)
b = (y + f(y)) / 2

Since b = (x + f(x)) / 2 and b = (y + f(y)) / 2, f is a reflection over the point b.

QED.