- #1
evry190
- 13
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0<x<y, prove x^n<y^n for n = natural numbers
I know its obvious, but I don't really know what to write to proove it...
I know its obvious, but I don't really know what to write to proove it...
jgens said:Induction seems like the best way to give a formal proof of the fact, but a sloppy and informal proof could show that [itex]x^n < x^{n-1}y < \dots < xy^{n-1} < y^n[/itex].
The inequality x^n < y^n means that when two numbers, x and y, are raised to the same power n, the result of y^n will always be greater than the result of x^n.
One way to prove the inequality x^n < y^n is by using mathematical induction. This involves showing that the inequality holds for a base case, and then proving that if it holds for a certain value of n, it also holds for the next value of n.
No, the inequality x^n < y^n is not always true for all values of x and y. It depends on the value of n and the relationship between x and y. For example, if n is a negative number, the inequality will be reversed and x^n > y^n.
Yes, there are exceptions to the inequality x^n < y^n. For example, if x and y are both negative numbers and n is an even number, then x^n < y^n may not hold true. This is because when a negative number is raised to an even power, the result becomes positive.
Proving the inequality x^n < y^n is important in mathematics because it helps us understand and compare the relationships between numbers raised to different powers. It is also a fundamental concept in algebra and calculus, and is used in many mathematical proofs and applications.