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Is this proof of $x<y iff x^n < y^n$ correct?
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Claim: $x<y iff x^n < y^n$ for $nin mathbb N$. Edit: $x,y>0$
Proof:
Since that which is to be proved is biconditional we must prove both that $x<y implies x^n<y^n$ and $x<y impliedby x^n<y^n$ are true.
First we prove that $x<y implies x^n<y^n$ is true by induction. It is trivial to show that this is true for the case $n=1$, and I have proved earlier that it is also true for the case $n=2$ (so we do not need to prove this here.
Let us assume that the statement holds for some $n=k$. We seek to show that it is also true for $n=k+1$.
We can say that $$x^k+1 - y^k+1 equiv (x-y)(x^k+x^k-1y +...+xy^k-1 + y^k)$$ and since $(x-y)<0$, and $(x^k+x^k-1y +...+xy^k-1 + y^k) > 0$, we see that $x^k+1 - y^k+1$ must be negative. Therefore we arrive at the desired result:
$$x^k+1 - y^k+1<0.$$
Thus for all $nin mathbb N$, $x<y implies x^n<y^n$.
Now we must prove that $x<y impliedby x^n<y^n$ is also true. If we write this statement in the contrapositive form, we find that:
$$(x<y impliedby x^n<y^n) iff (x^n<y^n implies x<y)$$
$$iff (y<x implies y^n<x^n).$$
Since $x$ and $y$ are arbitrary numbers, we have already proven this. $$tag*$blacksquare$$$
--
The above is my attempt at the proof, I would like for someone to confirm whether it is correct or not. For the line where I state the equivalence $x^k+1 - y^k+1 equiv (x-y)(x^k+x^k-1y +...+xy^k-1 + y^k)$, I must make clear that I have had to use this as a given and would appreciate if someone could point out why this is obvious or how to go about proving this, almost as a lemma for the proof. My final request is that if there is any problem with my actual proofwriting, I ask that you point it out (e.g is this a usual style, or is it unsual and hard to follow, etc).
Edit: It was pointed out that I forgot to include that $x$ and $y$ are positive, so I have included this.
proof-verification proof-writing
asked Jul 15 at 19:24
Benjamin
473314
 |Â
show 3 more comments
up vote
3
down vote
favorite
Claim: $x<y iff x^n < y^n$ for $nin mathbb N$. Edit: $x,y>0$
Proof:
Since that which is to be proved is biconditional we must prove both that $x<y implies x^n<y^n$ and $x<y impliedby x^n<y^n$ are true.
First we prove that $x<y implies x^n<y^n$ is true by induction. It is trivial to show that this is true for the case $n=1$, and I have proved earlier that it is also true for the case $n=2$ (so we do not need to prove this here.
Let us assume that the statement holds for some $n=k$. We seek to show that it is also true for $n=k+1$.
We can say that $$x^k+1 - y^k+1 equiv (x-y)(x^k+x^k-1y +...+xy^k-1 + y^k)$$ and since $(x-y)<0$, and $(x^k+x^k-1y +...+xy^k-1 + y^k) > 0$, we see that $x^k+1 - y^k+1$ must be negative. Therefore we arrive at the desired result:
$$x^k+1 - y^k+1<0.$$
Thus for all $nin mathbb N$, $x<y implies x^n<y^n$.
Now we must prove that $x<y impliedby x^n<y^n$ is also true. If we write this statement in the contrapositive form, we find that:
$$(x<y impliedby x^n<y^n) iff (x^n<y^n implies x<y)$$
$$iff (y<x implies y^n<x^n).$$
Since $x$ and $y$ are arbitrary numbers, we have already proven this. $$tag*$blacksquare$$$
--
The above is my attempt at the proof, I would like for someone to confirm whether it is correct or not. For the line where I state the equivalence $x^k+1 - y^k+1 equiv (x-y)(x^k+x^k-1y +...+xy^k-1 + y^k)$, I must make clear that I have had to use this as a given and would appreciate if someone could point out why this is obvious or how to go about proving this, almost as a lemma for the proof. My final request is that if there is any problem with my actual proofwriting, I ask that you point it out (e.g is this a usual style, or is it unsual and hard to follow, etc).
Edit: It was pointed out that I forgot to include that $x$ and $y$ are positive, so I have included this.
proof-verification proof-writing
asked Jul 15 at 19:24
Benjamin
473314
Are you assuming $x, y > 0$?
– Dave
Jul 15 at 19:26
1
You need the condition that $x$ and $y$ are positive.
– Cameron Williams
Jul 15 at 19:26
1
true by inductionAs posted, that's a direct proof, since the induction hypothesis is never used.
– dxiv
Jul 15 at 19:28
2
The negation of $x < y$ is $y leq x$, not $y < x$.
– Dave
Jul 15 at 19:29
To see why you need positivity, consider the general case. To argue that $x^2 < y^2$ from $x < y$, we would want to multiply on both sides by $x$ so that $x^2 < xy$, but since $x < y$, $xy < y^2$ by multiplying on both sides by $y$. Note that if you don't assume positivity, the inequalities can get jumbled. (Say nothing of what happens with $0$.)
– Cameron Williams
Jul 15 at 19:29
 |Â
show 3 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Claim: $x<y iff x^n < y^n$ for $nin mathbb N$. Edit: $x,y>0$
Proof:
Since that which is to be proved is biconditional we must prove both that $x<y implies x^n<y^n$ and $x<y impliedby x^n<y^n$ are true.
First we prove that $x<y implies x^n<y^n$ is true by induction. It is trivial to show that this is true for the case $n=1$, and I have proved earlier that it is also true for the case $n=2$ (so we do not need to prove this here.
Let us assume that the statement holds for some $n=k$. We seek to show that it is also true for $n=k+1$.
We can say that $$x^k+1 - y^k+1 equiv (x-y)(x^k+x^k-1y +...+xy^k-1 + y^k)$$ and since $(x-y)<0$, and $(x^k+x^k-1y +...+xy^k-1 + y^k) > 0$, we see that $x^k+1 - y^k+1$ must be negative. Therefore we arrive at the desired result:
$$x^k+1 - y^k+1<0.$$
Thus for all $nin mathbb N$, $x<y implies x^n<y^n$.
Now we must prove that $x<y impliedby x^n<y^n$ is also true. If we write this statement in the contrapositive form, we find that:
$$(x<y impliedby x^n<y^n) iff (x^n<y^n implies x<y)$$
$$iff (y<x implies y^n<x^n).$$
Since $x$ and $y$ are arbitrary numbers, we have already proven this. $$tag*$blacksquare$$$
--
The above is my attempt at the proof, I would like for someone to confirm whether it is correct or not. For the line where I state the equivalence $x^k+1 - y^k+1 equiv (x-y)(x^k+x^k-1y +...+xy^k-1 + y^k)$, I must make clear that I have had to use this as a given and would appreciate if someone could point out why this is obvious or how to go about proving this, almost as a lemma for the proof. My final request is that if there is any problem with my actual proofwriting, I ask that you point it out (e.g is this a usual style, or is it unsual and hard to follow, etc).
Edit: It was pointed out that I forgot to include that $x$ and $y$ are positive, so I have included this.
proof-verification proof-writing
asked Jul 15 at 19:24
Benjamin
473314
Claim: $x<y iff x^n < y^n$ for $nin mathbb N$. Edit: $x,y>0$
Proof:
Since that which is to be proved is biconditional we must prove both that $x<y implies x^n<y^n$ and $x<y impliedby x^n<y^n$ are true.
First we prove that $x<y implies x^n<y^n$ is true by induction. It is trivial to show that this is true for the case $n=1$, and I have proved earlier that it is also true for the case $n=2$ (so we do not need to prove this here.
Let us assume that the statement holds for some $n=k$. We seek to show that it is also true for $n=k+1$.
We can say that $$x^k+1 - y^k+1 equiv (x-y)(x^k+x^k-1y +...+xy^k-1 + y^k)$$ and since $(x-y)<0$, and $(x^k+x^k-1y +...+xy^k-1 + y^k) > 0$, we see that $x^k+1 - y^k+1$ must be negative. Therefore we arrive at the desired result:
$$x^k+1 - y^k+1<0.$$
Thus for all $nin mathbb N$, $x<y implies x^n<y^n$.
Now we must prove that $x<y impliedby x^n<y^n$ is also true. If we write this statement in the contrapositive form, we find that:
$$(x<y impliedby x^n<y^n) iff (x^n<y^n implies x<y)$$
$$iff (y<x implies y^n<x^n).$$
Since $x$ and $y$ are arbitrary numbers, we have already proven this. $$tag*$blacksquare$$$
--
The above is my attempt at the proof, I would like for someone to confirm whether it is correct or not. For the line where I state the equivalence $x^k+1 - y^k+1 equiv (x-y)(x^k+x^k-1y +...+xy^k-1 + y^k)$, I must make clear that I have had to use this as a given and would appreciate if someone could point out why this is obvious or how to go about proving this, almost as a lemma for the proof. My final request is that if there is any problem with my actual proofwriting, I ask that you point it out (e.g is this a usual style, or is it unsual and hard to follow, etc).
Edit: It was pointed out that I forgot to include that $x$ and $y$ are positive, so I have included this.
proof-verification proof-writing
asked Jul 15 at 19:24
Benjamin
473314
edited Jul 15 at 19:30
asked Jul 15 at 19:24
Benjamin
473314
asked Jul 15 at 19:24
Benjamin
473314
asked Jul 15 at 19:24
Benjamin
473314
473314
Are you assuming $x, y > 0$?
– Dave
Jul 15 at 19:26
1
You need the condition that $x$ and $y$ are positive.
– Cameron Williams
Jul 15 at 19:26
1
true by inductionAs posted, that's a direct proof, since the induction hypothesis is never used.
– dxiv
Jul 15 at 19:28
2
The negation of $x < y$ is $y leq x$, not $y < x$.
– Dave
Jul 15 at 19:29
To see why you need positivity, consider the general case. To argue that $x^2 < y^2$ from $x < y$, we would want to multiply on both sides by $x$ so that $x^2 < xy$, but since $x < y$, $xy < y^2$ by multiplying on both sides by $y$. Note that if you don't assume positivity, the inequalities can get jumbled. (Say nothing of what happens with $0$.)
– Cameron Williams
Jul 15 at 19:29
 |Â
show 3 more comments
Are you assuming $x, y > 0$?
– Dave
Jul 15 at 19:26
1
You need the condition that $x$ and $y$ are positive.
– Cameron Williams
Jul 15 at 19:26
1
true by inductionAs posted, that's a direct proof, since the induction hypothesis is never used.
– dxiv
Jul 15 at 19:28
2
The negation of $x < y$ is $y leq x$, not $y < x$.
– Dave
Jul 15 at 19:29
To see why you need positivity, consider the general case. To argue that $x^2 < y^2$ from $x < y$, we would want to multiply on both sides by $x$ so that $x^2 < xy$, but since $x < y$, $xy < y^2$ by multiplying on both sides by $y$. Note that if you don't assume positivity, the inequalities can get jumbled. (Say nothing of what happens with $0$.)
– Cameron Williams
Jul 15 at 19:29
Are you assuming $x, y > 0$?
– Dave
Jul 15 at 19:26
Are you assuming $x, y > 0$?
– Dave
Jul 15 at 19:26
1
1
You need the condition that $x$ and $y$ are positive.
– Cameron Williams
Jul 15 at 19:26
You need the condition that $x$ and $y$ are positive.
– Cameron Williams
Jul 15 at 19:26
1
1
true by induction As posted, that's a direct proof, since the induction hypothesis is never used.– dxiv
Jul 15 at 19:28
true by induction As posted, that's a direct proof, since the induction hypothesis is never used.– dxiv
Jul 15 at 19:28
2
2
The negation of $x < y$ is $y leq x$, not $y < x$.
– Dave
Jul 15 at 19:29
The negation of $x < y$ is $y leq x$, not $y < x$.
– Dave
Jul 15 at 19:29
To see why you need positivity, consider the general case. To argue that $x^2 < y^2$ from $x < y$, we would want to multiply on both sides by $x$ so that $x^2 < xy$, but since $x < y$, $xy < y^2$ by multiplying on both sides by $y$. Note that if you don't assume positivity, the inequalities can get jumbled. (Say nothing of what happens with $0$.)
– Cameron Williams
Jul 15 at 19:29
To see why you need positivity, consider the general case. To argue that $x^2 < y^2$ from $x < y$, we would want to multiply on both sides by $x$ so that $x^2 < xy$, but since $x < y$, $xy < y^2$ by multiplying on both sides by $y$. Note that if you don't assume positivity, the inequalities can get jumbled. (Say nothing of what happens with $0$.)
– Cameron Williams
Jul 15 at 19:29
 |Â
show 3 more comments
1 Answer
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up vote
1
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accepted
Hint:
For $n=1$, $x<yiff x<y$ obviously holds.
Now assume that for some $n$,
$$x<yiff x^n<y^n.$$
Then by the rule of multiplication of inequalities,
$$x<yland x^n<y^nimplies x^n+1<y^n+1$$
so that
$$x<yimplies x^n<y^nimplies x^n+1<y^n+1.$$
Now try the contrapositive.
answered Jul 15 at 19:57
Yves Daoust
111k665204
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint:
For $n=1$, $x<yiff x<y$ obviously holds.
Now assume that for some $n$,
$$x<yiff x^n<y^n.$$
Then by the rule of multiplication of inequalities,
$$x<yland x^n<y^nimplies x^n+1<y^n+1$$
so that
$$x<yimplies x^n<y^nimplies x^n+1<y^n+1.$$
Now try the contrapositive.
answered Jul 15 at 19:57
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
accepted
Hint:
For $n=1$, $x<yiff x<y$ obviously holds.
Now assume that for some $n$,
$$x<yiff x^n<y^n.$$
Then by the rule of multiplication of inequalities,
$$x<yland x^n<y^nimplies x^n+1<y^n+1$$
so that
$$x<yimplies x^n<y^nimplies x^n+1<y^n+1.$$
Now try the contrapositive.
answered Jul 15 at 19:57
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint:
For $n=1$, $x<yiff x<y$ obviously holds.
Now assume that for some $n$,
$$x<yiff x^n<y^n.$$
Then by the rule of multiplication of inequalities,
$$x<yland x^n<y^nimplies x^n+1<y^n+1$$
so that
$$x<yimplies x^n<y^nimplies x^n+1<y^n+1.$$
Now try the contrapositive.
answered Jul 15 at 19:57
Yves Daoust
111k665204
Hint:
For $n=1$, $x<yiff x<y$ obviously holds.
Now assume that for some $n$,
$$x<yiff x^n<y^n.$$
Then by the rule of multiplication of inequalities,
$$x<yland x^n<y^nimplies x^n+1<y^n+1$$
so that
$$x<yimplies x^n<y^nimplies x^n+1<y^n+1.$$
Now try the contrapositive.
answered Jul 15 at 19:57
Yves Daoust
111k665204
answered Jul 15 at 19:57
Yves Daoust
111k665204
answered Jul 15 at 19:57
Yves Daoust
111k665204
answered Jul 15 at 19:57
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
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Are you assuming $x, y > 0$?
– Dave
Jul 15 at 19:26
1
You need the condition that $x$ and $y$ are positive.
– Cameron Williams
Jul 15 at 19:26
1
true by inductionAs posted, that's a direct proof, since the induction hypothesis is never used.– dxiv
Jul 15 at 19:28
2
The negation of $x < y$ is $y leq x$, not $y < x$.
– Dave
Jul 15 at 19:29
To see why you need positivity, consider the general case. To argue that $x^2 < y^2$ from $x < y$, we would want to multiply on both sides by $x$ so that $x^2 < xy$, but since $x < y$, $xy < y^2$ by multiplying on both sides by $y$. Note that if you don't assume positivity, the inequalities can get jumbled. (Say nothing of what happens with $0$.)
– Cameron Williams
Jul 15 at 19:29