Stephanie M. answered • 07/10/15
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Since this is an "if an only if" statement, you'll essentially need two proofs:
1. PROVE THAT IF F IS A TRANSLATION, THEN F IS INCREASING
Assume f is a translation. Let x and y be two real numbers where x < y. We'd like to prove that f(x) < f(y) (and thus that f is increasing). A translation follows the rule (a --> a+b) for some real numbers a and b. So, f(x) = x+b and f(y) = y+b. Since x < y, then x+b < y+b (by the addition property of inequality) and so f(x) < f(y). Thus, f is increasing.
2. PROVE THAT IF F IS INCREASING, THEN F IS A TRANSLATION
Assume f is increasing. Then for x < y, f(x) < f(y). Since isometries preserve distance by definition, d(x, y) = d(f(x), f(y)). This means that x - y = f(x) - f(y). Let f(x) = x + b. Then:
f(x) - b = x
f(x) - b - y = f(x) - f(y) (SUBSTITUTION)
-b - y = -f(y)
b + y = f(y)
f(y) = y + b
Thus, f(x) = x + b and f(y) = y + b. This is the same as a translation with the rule (a --> a+b) for some real numbers a and b. Thus, f is a translation.
QED.
This ought to give you a good idea of how to write your own version of the proof. In general, you'll just want to use the math definitions of various terms (isometry means d(x, y) = d(f(x), f(y)), translation means a --> a+b, increasing means x < y implies f(x) < f(y), etc.) to move from the statement (IF) to the conclusion (THEN).
Stephanie M.
tutor
So, 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.
Does that make sense?
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07/10/15
Megan O.
07/10/15