Proving x^n<y^n with Michael Spivak's Calculus

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SUMMARY

The discussion focuses on proving the inequality x^n < y^n for n = 1, 2, 3, ... given that 0 ≤ x < y. The user successfully demonstrated the case for n = 2 but struggled with generalizing the proof for all n. A suggested approach involves using mathematical induction, where the relation x^{n+1} = x(x^n) < x(y^n) < y(y^n) = y^{n+1} is established. Additionally, the user proposed a direct proof using the factorization x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + ... + y^{n-1}), which also supports the inequality.

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Students of calculus, particularly those studying Michael Spivak's Calculus, educators teaching mathematical proofs, and anyone interested in mastering inequalities and induction techniques.

Einherjer
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Just started on Michael Spivaks Calculus, going fine so far, but generalizing to n always messes me up.

Homework Statement


prove that If 0≤x<y, then x^n<y^n, n=1,2,3,...

The Attempt at a Solution


I have managed to to prove that x^2<y^2. But I'm i don't quite get how to generalize to n.

First time posting a question here, so tell me if I've done something wrong.
 
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Are you using a proof by induction? Then [itex]x^{n+1}= x(x^n)< x(y^n)< y(y^n)= y^{n+1}[/itex]

But I would think that [itex]x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ x^{n-3}y^2+[/itex][itex]\cdot\cdot\cdot+ x^2y^{n-3}+ xy^{n-2}+ y^{n-1})[/itex] would prove it directly.
 

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