Prove that the equation $x^2=y$ has a unique solution
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If $y$ is a real number greater than zero, there is only one real number $x$ greater than zero such that:
$$x^2=y$$
The similar question For every real $x>0$ and every integer $n>0$, there is one and only one real $y>0$ such that $y^n=x$ has been asked, but the answers reach a depth greater than I'm comfortable with for what seems to be a more basic proof.
My strategy is:
Suppose that there are positive numbers $a$ and $b$ each of whose square is $c$, then:
$$0=a^2-b^2=(a-b)(a+b)$$
Since since $a+b$ is greater than $0$, it follows that $a=b$. q.e.d.
Is this a good strategy/does it accurately prove the proof? The textbook I've been reading from has mentioned this general proposition twice and mentioned proving it, but later on, however I felt it was strange as it seems we already have the tools to prove it.
real-analysis proof-verification
edited Jun 4 at 19:34
Emilio Novati
50.8k43472
asked Jun 4 at 19:23
Bad at algebra and proofs
16411
add a comment |
up vote
11
down vote
favorite
If $y$ is a real number greater than zero, there is only one real number $x$ greater than zero such that:
$$x^2=y$$
The similar question For every real $x>0$ and every integer $n>0$, there is one and only one real $y>0$ such that $y^n=x$ has been asked, but the answers reach a depth greater than I'm comfortable with for what seems to be a more basic proof.
My strategy is:
Suppose that there are positive numbers $a$ and $b$ each of whose square is $c$, then:
$$0=a^2-b^2=(a-b)(a+b)$$
Since since $a+b$ is greater than $0$, it follows that $a=b$. q.e.d.
Is this a good strategy/does it accurately prove the proof? The textbook I've been reading from has mentioned this general proposition twice and mentioned proving it, but later on, however I felt it was strange as it seems we already have the tools to prove it.
real-analysis proof-verification
edited Jun 4 at 19:34
Emilio Novati
50.8k43472
asked Jun 4 at 19:23
Bad at algebra and proofs
16411
4
Yes, your proof is correct, and yes, it can be extended to $n>2.$ Can you factor $a^n-b^n?$ Hint: $a-b$ is one of the factors.
– saulspatz
Jun 4 at 19:29
4
Your proof is correct, but say that if $a>0$ and $b>0$ exist such that $a^2=b^2=y$ than $a=b$, but the difficult task is to prove that $a$ ( or $b$) really exist. This is true in the realm or real numbers, but not in the realm of rational.
– Emilio Novati
Jun 4 at 19:38
add a comment |
up vote
11
down vote
favorite
up vote
11
down vote
favorite
If $y$ is a real number greater than zero, there is only one real number $x$ greater than zero such that:
$$x^2=y$$
The similar question For every real $x>0$ and every integer $n>0$, there is one and only one real $y>0$ such that $y^n=x$ has been asked, but the answers reach a depth greater than I'm comfortable with for what seems to be a more basic proof.
My strategy is:
Suppose that there are positive numbers $a$ and $b$ each of whose square is $c$, then:
$$0=a^2-b^2=(a-b)(a+b)$$
Since since $a+b$ is greater than $0$, it follows that $a=b$. q.e.d.
Is this a good strategy/does it accurately prove the proof? The textbook I've been reading from has mentioned this general proposition twice and mentioned proving it, but later on, however I felt it was strange as it seems we already have the tools to prove it.
real-analysis proof-verification
edited Jun 4 at 19:34
Emilio Novati
50.8k43472
asked Jun 4 at 19:23
Bad at algebra and proofs
16411
If $y$ is a real number greater than zero, there is only one real number $x$ greater than zero such that:
$$x^2=y$$
The similar question For every real $x>0$ and every integer $n>0$, there is one and only one real $y>0$ such that $y^n=x$ has been asked, but the answers reach a depth greater than I'm comfortable with for what seems to be a more basic proof.
My strategy is:
Suppose that there are positive numbers $a$ and $b$ each of whose square is $c$, then:
$$0=a^2-b^2=(a-b)(a+b)$$
Since since $a+b$ is greater than $0$, it follows that $a=b$. q.e.d.
Is this a good strategy/does it accurately prove the proof? The textbook I've been reading from has mentioned this general proposition twice and mentioned proving it, but later on, however I felt it was strange as it seems we already have the tools to prove it.
real-analysis proof-verification
real-analysis proof-verification
edited Jun 4 at 19:34
Emilio Novati
50.8k43472
asked Jun 4 at 19:23
Bad at algebra and proofs
16411
edited Jun 4 at 19:34
Emilio Novati
50.8k43472
asked Jun 4 at 19:23
Bad at algebra and proofs
16411
edited Jun 4 at 19:34
Emilio Novati
50.8k43472
edited Jun 4 at 19:34
Emilio Novati
50.8k43472
edited Jun 4 at 19:34
Emilio Novati
50.8k43472
50.8k43472
asked Jun 4 at 19:23
Bad at algebra and proofs
16411
asked Jun 4 at 19:23
Bad at algebra and proofs
16411
asked Jun 4 at 19:23
Bad at algebra and proofs
16411
16411
4
Yes, your proof is correct, and yes, it can be extended to $n>2.$ Can you factor $a^n-b^n?$ Hint: $a-b$ is one of the factors.
– saulspatz
Jun 4 at 19:29
4
Your proof is correct, but say that if $a>0$ and $b>0$ exist such that $a^2=b^2=y$ than $a=b$, but the difficult task is to prove that $a$ ( or $b$) really exist. This is true in the realm or real numbers, but not in the realm of rational.
– Emilio Novati
Jun 4 at 19:38
add a comment |
4
Yes, your proof is correct, and yes, it can be extended to $n>2.$ Can you factor $a^n-b^n?$ Hint: $a-b$ is one of the factors.
– saulspatz
Jun 4 at 19:29
4
Your proof is correct, but say that if $a>0$ and $b>0$ exist such that $a^2=b^2=y$ than $a=b$, but the difficult task is to prove that $a$ ( or $b$) really exist. This is true in the realm or real numbers, but not in the realm of rational.
– Emilio Novati
Jun 4 at 19:38
4
4
Yes, your proof is correct, and yes, it can be extended to $n>2.$ Can you factor $a^n-b^n?$ Hint: $a-b$ is one of the factors.
– saulspatz
Jun 4 at 19:29
Yes, your proof is correct, and yes, it can be extended to $n>2.$ Can you factor $a^n-b^n?$ Hint: $a-b$ is one of the factors.
– saulspatz
Jun 4 at 19:29
4
4
Your proof is correct, but say that if $a>0$ and $b>0$ exist such that $a^2=b^2=y$ than $a=b$, but the difficult task is to prove that $a$ ( or $b$) really exist. This is true in the realm or real numbers, but not in the realm of rational.
– Emilio Novati
Jun 4 at 19:38
Your proof is correct, but say that if $a>0$ and $b>0$ exist such that $a^2=b^2=y$ than $a=b$, but the difficult task is to prove that $a$ ( or $b$) really exist. This is true in the realm or real numbers, but not in the realm of rational.
– Emilio Novati
Jun 4 at 19:38
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
What you have shown is that given a real number $y > 0$ there is at most one $x > 0$ such that $x^2 = y$. You are yet to show that such an $x$ exists for every $y > 0$. For this you will need to use the completeness of the reals somewhere, as the answers to the linked question indicate. For instance, your proof works just as well over the rationals: as per your proof, for every rational $y > 0$ there is at most one rational $x > 0$ such that $x^2 = y$. But it is also possible that there is no such rational $x$ for some values of $y$ (for instance, take $y = 2$).
So, your strategy covers the uniqueness of the square root, whereas the existence of the square root still remains to be proved.
answered Nov 17 at 7:27
Brahadeesh
5,78441957
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
What you have shown is that given a real number $y > 0$ there is at most one $x > 0$ such that $x^2 = y$. You are yet to show that such an $x$ exists for every $y > 0$. For this you will need to use the completeness of the reals somewhere, as the answers to the linked question indicate. For instance, your proof works just as well over the rationals: as per your proof, for every rational $y > 0$ there is at most one rational $x > 0$ such that $x^2 = y$. But it is also possible that there is no such rational $x$ for some values of $y$ (for instance, take $y = 2$).
So, your strategy covers the uniqueness of the square root, whereas the existence of the square root still remains to be proved.
answered Nov 17 at 7:27
Brahadeesh
5,78441957
add a comment |
up vote
2
down vote
What you have shown is that given a real number $y > 0$ there is at most one $x > 0$ such that $x^2 = y$. You are yet to show that such an $x$ exists for every $y > 0$. For this you will need to use the completeness of the reals somewhere, as the answers to the linked question indicate. For instance, your proof works just as well over the rationals: as per your proof, for every rational $y > 0$ there is at most one rational $x > 0$ such that $x^2 = y$. But it is also possible that there is no such rational $x$ for some values of $y$ (for instance, take $y = 2$).
So, your strategy covers the uniqueness of the square root, whereas the existence of the square root still remains to be proved.
answered Nov 17 at 7:27
Brahadeesh
5,78441957
add a comment |
up vote
2
down vote
up vote
2
down vote
What you have shown is that given a real number $y > 0$ there is at most one $x > 0$ such that $x^2 = y$. You are yet to show that such an $x$ exists for every $y > 0$. For this you will need to use the completeness of the reals somewhere, as the answers to the linked question indicate. For instance, your proof works just as well over the rationals: as per your proof, for every rational $y > 0$ there is at most one rational $x > 0$ such that $x^2 = y$. But it is also possible that there is no such rational $x$ for some values of $y$ (for instance, take $y = 2$).
So, your strategy covers the uniqueness of the square root, whereas the existence of the square root still remains to be proved.
answered Nov 17 at 7:27
Brahadeesh
5,78441957
What you have shown is that given a real number $y > 0$ there is at most one $x > 0$ such that $x^2 = y$. You are yet to show that such an $x$ exists for every $y > 0$. For this you will need to use the completeness of the reals somewhere, as the answers to the linked question indicate. For instance, your proof works just as well over the rationals: as per your proof, for every rational $y > 0$ there is at most one rational $x > 0$ such that $x^2 = y$. But it is also possible that there is no such rational $x$ for some values of $y$ (for instance, take $y = 2$).
So, your strategy covers the uniqueness of the square root, whereas the existence of the square root still remains to be proved.
answered Nov 17 at 7:27
Brahadeesh
5,78441957
answered Nov 17 at 7:27
Brahadeesh
5,78441957
answered Nov 17 at 7:27
Brahadeesh
5,78441957
answered Nov 17 at 7:27
Brahadeesh
5,78441957
5,78441957
add a comment |
add a comment |
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Yes, your proof is correct, and yes, it can be extended to $n>2.$ Can you factor $a^n-b^n?$ Hint: $a-b$ is one of the factors.
– saulspatz
Jun 4 at 19:29
4
Your proof is correct, but say that if $a>0$ and $b>0$ exist such that $a^2=b^2=y$ than $a=b$, but the difficult task is to prove that $a$ ( or $b$) really exist. This is true in the realm or real numbers, but not in the realm of rational.
– Emilio Novati
Jun 4 at 19:38