Calculus Exercises (HELP-SOS)

b!tcoin

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Oct 7, 2019
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Hello guys.
First of all good job on making this awesome forum.
Secondly i'd like to ask for some help regarding my calculus exercises.
Here it is:

Prove by induction that |x1 + x2 + · · · + xn(little n)| ≤ |x1| + |x2| + · · · + |xn(little n)| for any n ∈ N and any numbers x1, . . . , xn(little n) ∈ Q.



AND


(a) Use the ordering axioms to show that x 2 < y2 for any x, y ∈ Q with 0 < x < y. (
b) Use the previous result to prove by induction that x^n < y^n for any n ∈ N and any x, y ∈ Q with 0 < x < y. ( I Have solved A, can someone help with b? thanks
 
What is the difficulty? Do you know how proof by induction works? Can you do the first step?
 
yes,in we firstly state the proposition in mathematical notation and then take base case n=1. but those | | are confusing me
 
It is always difficult to help with proofs because we do not know exactly what set of axioms you are using and what theorems you have previously proved

[MATH]x < y \iff \exists \ d \text { such that } y - x = d > 0.[/MATH]
We are given that 0 < x < y. Define n as a positive integer and

[MATH]z_n = y^n - x^n.[/MATH]
[MATH]z_n > 0 \implies x^n < y^n.[/MATH]
It is easy to show that z_1 > 0. How?

If n > 1

[MATH]z^n = y^n - x^n = (y - x) * WHAT?[/MATH]
 
yes for n=1 |x1| is lower or equal to |x1|
so now imma take the n+1 case
 
yes for n=1 |x1| is lower or equal to |x1|
so now imma take the n+1 case
If we know that k=1Jxkk=1Jxk\displaystyle \left| {\sum\limits_{k = 1}^J {{x_k}} } \right| \leqslant \sum\limits_{k = 1}^J {\left| {{x_k}} \right|}
Then define S=k=1Jxk\displaystyle S=\sum\limits_{k = 1}^J {{{x_k}} } What can we say about S+xJ+1 ?\displaystyle \left|S+x_{J+1}\right|\le~?
 
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