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I was doing Analysis when I came across with this problem

It reads that : Proof that if x<y , therefore x^n<y^n

Could anyone help me out with this ?

Thanks

Gary L

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- Thread starter garyljc
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- #1

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I was doing Analysis when I came across with this problem

It reads that : Proof that if x<y , therefore x^n<y^n

Could anyone help me out with this ?

Thanks

Gary L

- #2

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Watch out for negative signs

- #3

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This is false. For example, x=-2, y=1, n=2

- #4

CompuChip

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Suppose that

Somehow, you will have to put the assumption in there again, and there is as far as I see only one obvious way to do it... how would you rewrite, for example, the left hand side of the inequality?

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You may use the axiom that if x>y, then xz>yz, for any z>0

- #6

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Or should I actually do it by case analysis , where by I consider both positive and negative ?

- #7

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Suppose that0 <x < y, and we have annintegersuch that x^n < y^n. Prove that x^(n + 1) < y^(n + 1).

Somehow, you will have to put the assumption in there again, and there is as far as I see only one obvious way to do it... how would you rewrite, for example, the left hand side of the inequality?

I'll just simplify it to become x^n.x < y^n.y

is that it ?

- #8

CompuChip

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- #9

HallsofIvy

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Have you already proved that: it a< b and x< y, then ax< by?I'll just simplify it to become x^n.x < y^n.y

is that it ?

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Have you already proved that: if a< b and 0< x< y, then ax< by?

is this the first very step i have to do in order to proceed ?

correct me if i'm wrong , but without this step , does it mean this equation might not hold ?

- #11

HallsofIvy

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I assume that your induction step is "if x^{n}< y^{n}, then (x^{n})x< (y^{n})y so x^{n+1}< y^{n+1}". In order to go from the second to the third inequality, you have to know that "if a< b and 0< x< y, then ax< by" which is NOT exactly the same as the order axiom "if a< b and 0< x then ax< bx".

Of course, it's easy to prove that for all positive numbers which is true here:

If 0< a< b, and 0< x< y, then, first, ax< bx (multiplying both sides of a< b by the positive number x) and, second, bx< by (multiplying both sides of x< y by the positive number b). The result follows from transitivity.

Of course, it's easy to prove that for all positive numbers which is true here:

If 0< a< b, and 0< x< y, then, first, ax< bx (multiplying both sides of a< b by the positive number x) and, second, bx< by (multiplying both sides of x< y by the positive number b). The result follows from transitivity.

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got it =)

thanks

- #13

CompuChip

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Can you show us the full proof? Just to check that you haven't forgotten anything.

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I assume that your induction step is "if x^{n[/sub]< yn, then (xn)x< (yn)y so xn+1< yn+1". In order to go from the second to the third inequality, you have to know that "if a< b and 0< x< y, then ax< by" which is NOT exactly the same as the order axiom "if a< b and 0< x then ax< bx". Of course, it's easy to prove that for all positive numbers which is true here: If 0< a< b, and 0< x< y, then, first, ax< bx (multiplying both sides of a< b by the positive number x) and, second, bx< by (multiplying both sides of x< y by the positive number b). The result follows from transitivity.}

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Can you show us the full proof? Just to check that you haven't forgotten anything.

my steps are actually discussed throughout the thread ...

but now , i got another similar proof ... but it reads something like contra positive

whereby i have to proof this similar statement from the RHS instead of LHS

is there anything additional that i need to consider ?

- #16

CompuChip

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A, thenB" then how would you prove this from the contrapositive.

so we must prove that not A then not B ?

- #18

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x<y , we want to prove that x^n<y^n

assuming x^n<y^n , x^(n+1)<y^(n+1) must be also true

0<x<y

therefore x^(n+1)<y^(n+1)

is x^n . x < y^n . y

and since x<y

therefore it is true for all values of n #

is that correct ?

- #19

HallsofIvy

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As noted before,

x<y , we want to prove that x^n<y^n

assuming x^n<y^n , x^(n+1)<y^(n+1) must be also true

0<x<y

therefore x^(n+1)<y^(n+1)

is x^n . x < y^n . y

and since x<y

therefore it is true for all values of n #

is that correct ?

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so we must prove that not A then not B ?

If A => B is your original implication, then the contrapositive would be ~B => ~A. ~A => ~B would be the inverse of the original implication and the converse of the contrapositive.

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As noted before,haveyou already proved "if x< y and 0< a< b, then ax< by"?

halls i dont get it

but isn't that given by the question ?

- #22

HallsofIvy

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"but now , i got another similar proof ... but it reads something like contra positive

whereby i have to proof this similar statement from the RHS instead of LHS

is there anything additional that i need to consider ? "

I thought you were still on the first question!

The contrapositive of "If A then B" is "If

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lol ...

but halls ... even for the 1st question

do i need to prove the axiom ?

but halls ... even for the 1st question

do i need to prove the axiom ?

- #24

CompuChip

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But I wouldn't say it is an axiom, just a theorem. And I don't think that it is trivial (if it is, then the entire question is rather trivial as it immediately follows from the axiom). I think the following is an axiom,

If 0 < a < b then for any x > 0, a x < b x.

Note how there is just an If 0 < a < b then for any y > x > 0, a x < b y.

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so could you take a look at what i've come up for the first proof

is it sufficient to proof it ?

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