Need help with this induction proof

[garyljc]

Hi all,
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

 

[maze]

Watch out for negative signs

 

[uman]

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

 

[CompuChip]

So the main part of the proof is (note the extra assumptions I had to make, in bold):
Suppose that 0 < x < y, and we have an n integer such 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?

 

[Kurret]

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

 

[garyljc]

Should I state that in order for this to be true both x and y have to be positive ?

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

 

[garyljc]

So the main part of the proof is (note the extra assumptions I had to make, in bold):
Suppose that 0 < x < y, and we have an n integer such 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 ?

 

[CompuChip]

Yes, you could do that (though it's more like the reverse of a simplification, as you're writing it in a more difficult way -- though more convenient in this case). But you haven't proven why that equation holds yet. Remember, all you know is that x < y and that x^n < y^n. Now try to combine that with the "simplification" to prove x^(n + 1) < y^(n + 1). Also look at Kurret's hint, it's quite useful.

 

[HallsofIvy]

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

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

 

[garyljc]

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 ?

 

[HallsofIvy]

I assume that your induction step is "if xⁿ< 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.

 

[garyljc]

Yes, you could do that (though it's more like the reverse of a simplification, as you're writing it in a more difficult way -- though more convenient in this case). But you haven't proven why that equation holds yet. Remember, all you know is that x < y and that x^n < y^n. Now try to combine that with the "simplification" to prove x^(n + 1) < y^(n + 1). Also look at Kurret's hint, it's quite useful.

got it =)
thanks

 

[CompuChip]

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

 

[garyljc]

I assume that your induction step is "if x<sup>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.</sup>

^{thanks halls I've got it now}

 

[garyljc]

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 ?

 

[CompuChip]

Yes, you should consider the way a contrapositive proof works. If the statement is: "If A, then B" then how would you prove this from the contrapositive.

 

[garyljc]

Yes, you should consider the way a contrapositive proof works. If the statement is: "If A, then B" then how would you prove this from the contrapositive.

so we must prove that not A then not B ?

 

[garyljc]

this is what i came up with for the first step
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 ?

 

[HallsofIvy]

this is what i came up with for the first step
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 ?

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

 

[The|M|onster]

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.

 

[garyljc]

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

halls i don't get it
but isn't that given by the question ?

 

[HallsofIvy]

Sorry, I hadn't realized that you had introduced a completely new question:
"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 not B then not A" and is equivalent to the original statement- proving one proves the other.

 

[garyljc]

lol ...
but halls ... even for the 1st question
do i need to prove the axiom ?

 

[CompuChip]

You call it an axiom, which is by definition something you don't prove.
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 x on both sides of the inequality, whereas you want to prove that

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

 

[garyljc]

compu chip thanks for the advice
so could you take a look at what I've come up for the first proof
is it sufficient to proof it ?

 

[garyljc]

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

is anyone here ?
help me
i've post the proof

halls , i was wondering if i should just print this "if x< y and 0< a< b, then ax< by" since it's rather trivial
it's a one line proof

 

[garyljc]

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

this is very trivial
do we have to proof it ?
if so , how do we ?

 

[CompuChip]

Well, how trivial it is depends on how precise you want to get. I could also say that what you want to prove is trivial and you don't need to prove it, but apparently you think otherwise
Currently, your proof is globally of the form:
"Suppose <induction hypothesis>, then <what we want to prove> is true so we have proven <what we want to prove>".

So you can assume this axiom (as said before)

If x<y, then xz<yz, for any z>0​

Note that on both sides you are multiplying by the same number. So the axiom does not say that:

If 0<x<y, then x^2<y^2​

because z would have to be equal to x and y at the same time. So first try to find the missing step(s) here.

 

[HallsofIvy]

is anyone here ?
help me
i've post the proof

halls , i was wondering if i should just print this "if x< y and 0< a< b, then ax< by" since it's rather trivial
it's a one line proof

this is very trivial
do we have to proof it ?
if so , how do we ?

You've said twice now that it is "trivial" and once that it is "a one line proof". Why are you now asking HOW to prove it?

Actually, the statement, as given here, is NOT true. For example, if a= 1, b= 2, x= -3, y= -2 then x< y, 0<a< b but ax= -3, by= -4 so ax is NOT less than by. You need at least y> 0 also. In that case, since x< y and a> 0, ax< ay. Since a< b and y> 0, ay< by. From ax< ay< by, ax< by.