Basic question about proving uniqueness The Next CEO of Stack OverflowProving every positive...
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Basic question about proving uniqueness
The Next CEO of Stack OverflowProving every positive natural number has unique predecessorProving there is a unique binary operation we call multiplicationUniqueness of modular multiplicative inversechange of order in a double seriesShowing that the open ball is homeomorphic to $mathbb{R}^n$Question about required rigour with intro real analysis textUniqueness of a Limit epsilon divided by 2?Fundamental Existence-Uniqueness Theorem of Nonautonomous SystemsProving non-differentiability in a basic senseAnother question about proving Lebesgue Decompositionquestion about limitAmbiguity in Question!Proof of uniqueness of positive cubes
$begingroup$
I just started real analysis. I don't have a background in proofs or logic, simply calculus. So I'm trying to learn more about proofs--so forgive the basic question, please.
How do you go about proving this theorem: If $a$ and $b$ are any numbers, then there is one and only one number $x$ such that $a + x=b$. This number is given by $x=b+(-a)$.
First part of the proof I understand--it's simple. We simply use some axioms and do the following:
$a+b-a=b$, which of course is true. This is just true from plugging in the $b+(-a)$ for $x$.
But how about the uniqueness issue? In my text, it says:
$$(a+x)+(-a)=b+(-a)
x=b+(-a)$$
How does this prove uniqueness? Is this equation essentially saying that the ONLY possible value of $x$ is expressed by $b+(-a)$?
How would I have known to set this up as part of the proof if I was asked to prove something is unique?
Sorry for the basic question--have to start somewhere.
real-analysis
asked Dec 23 '10 at 22:54
rstecklyrsteckly
12314
$endgroup$
add a comment |
$begingroup$
I just started real analysis. I don't have a background in proofs or logic, simply calculus. So I'm trying to learn more about proofs--so forgive the basic question, please.
How do you go about proving this theorem: If $a$ and $b$ are any numbers, then there is one and only one number $x$ such that $a + x=b$. This number is given by $x=b+(-a)$.
First part of the proof I understand--it's simple. We simply use some axioms and do the following:
$a+b-a=b$, which of course is true. This is just true from plugging in the $b+(-a)$ for $x$.
But how about the uniqueness issue? In my text, it says:
$$(a+x)+(-a)=b+(-a)
x=b+(-a)$$
How does this prove uniqueness? Is this equation essentially saying that the ONLY possible value of $x$ is expressed by $b+(-a)$?
How would I have known to set this up as part of the proof if I was asked to prove something is unique?
Sorry for the basic question--have to start somewhere.
real-analysis
asked Dec 23 '10 at 22:54
rstecklyrsteckly
12314
$endgroup$
$begingroup$
Yes, it is essentially saying that the only possible value is b+(-a).
$endgroup$
– Aryabhata
Dec 23 '10 at 23:04
$begingroup$
Relevant SMBC comic (I think)
$endgroup$
– Nick T
Jan 5 '14 at 19:21
add a comment |
$begingroup$
I just started real analysis. I don't have a background in proofs or logic, simply calculus. So I'm trying to learn more about proofs--so forgive the basic question, please.
How do you go about proving this theorem: If $a$ and $b$ are any numbers, then there is one and only one number $x$ such that $a + x=b$. This number is given by $x=b+(-a)$.
First part of the proof I understand--it's simple. We simply use some axioms and do the following:
$a+b-a=b$, which of course is true. This is just true from plugging in the $b+(-a)$ for $x$.
But how about the uniqueness issue? In my text, it says:
$$(a+x)+(-a)=b+(-a)
x=b+(-a)$$
How does this prove uniqueness? Is this equation essentially saying that the ONLY possible value of $x$ is expressed by $b+(-a)$?
How would I have known to set this up as part of the proof if I was asked to prove something is unique?
Sorry for the basic question--have to start somewhere.
real-analysis
asked Dec 23 '10 at 22:54
rstecklyrsteckly
12314
$endgroup$
I just started real analysis. I don't have a background in proofs or logic, simply calculus. So I'm trying to learn more about proofs--so forgive the basic question, please.
How do you go about proving this theorem: If $a$ and $b$ are any numbers, then there is one and only one number $x$ such that $a + x=b$. This number is given by $x=b+(-a)$.
First part of the proof I understand--it's simple. We simply use some axioms and do the following:
$a+b-a=b$, which of course is true. This is just true from plugging in the $b+(-a)$ for $x$.
But how about the uniqueness issue? In my text, it says:
$$(a+x)+(-a)=b+(-a)
x=b+(-a)$$
How does this prove uniqueness? Is this equation essentially saying that the ONLY possible value of $x$ is expressed by $b+(-a)$?
How would I have known to set this up as part of the proof if I was asked to prove something is unique?
Sorry for the basic question--have to start somewhere.
real-analysis
real-analysis
asked Dec 23 '10 at 22:54
rstecklyrsteckly
12314
asked Dec 23 '10 at 22:54
rstecklyrsteckly
12314
edited Mar 9 '17 at 19:41
user409521
asked Dec 23 '10 at 22:54
rstecklyrsteckly
12314
asked Dec 23 '10 at 22:54
rstecklyrsteckly
12314
asked Dec 23 '10 at 22:54
rstecklyrsteckly
12314
12314
$begingroup$
Yes, it is essentially saying that the only possible value is b+(-a).
$endgroup$
– Aryabhata
Dec 23 '10 at 23:04
$begingroup$
Relevant SMBC comic (I think)
$endgroup$
– Nick T
Jan 5 '14 at 19:21
add a comment |
$begingroup$
Yes, it is essentially saying that the only possible value is b+(-a).
$endgroup$
– Aryabhata
Dec 23 '10 at 23:04
$begingroup$
Relevant SMBC comic (I think)
$endgroup$
– Nick T
Jan 5 '14 at 19:21
$begingroup$
Yes, it is essentially saying that the only possible value is b+(-a).
$endgroup$
– Aryabhata
Dec 23 '10 at 23:04
$begingroup$
Yes, it is essentially saying that the only possible value is b+(-a).
$endgroup$
– Aryabhata
Dec 23 '10 at 23:04
$begingroup$
Relevant SMBC comic (I think)
$endgroup$
– Nick T
Jan 5 '14 at 19:21
$begingroup$
Relevant SMBC comic (I think)
$endgroup$
– Nick T
Jan 5 '14 at 19:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Usually when you want to prove something is unique you start by assuming there are two things that satisfy the given property, and then you show that they are the same. In this case, if you want to prove that there is exactly one $x$ such that $a + x = b$, you could start by assuming there are two numbers, say $x$ and $y$ that satisfy the identity, that is
$$a + x = b quad text{and} quad a + y = b$$
but then this implies that $a + x = a + y$ since both are equal to $b$. Now you can add $-a$ to both sides of the equation to get
$$-a + (a + x) = -a + (a + y) implies (-a + a) + x = (-a + a) + y $$
$$implies 0 + x = 0 + y implies x = y$$
This takes care of the uniqueness of the number and since you already know one such $x$ then you're done. I suppose you're doing this in the real number field and that you know which axioms take place in the argument. Hope this helps a little.
answered Dec 23 '10 at 23:10
Adrián BarqueroAdrián Barquero
10.8k23982
$endgroup$
$begingroup$
So proof by contradiction? But what about the given argument, is it really air tight?
$endgroup$
– rsteckly
Dec 23 '10 at 23:13
1
$begingroup$
Yes, it is correct. What I did is to assume two numbers $x, y$ satisfy the identity and proved that $x = y$. What your text does is to show what $x$ should be, so it is another way of proving uniqueness.
$endgroup$
– Adrián Barquero
Dec 23 '10 at 23:16
$begingroup$
án: But it's redundant to give such a uniqueness proof - see my answer.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 19:57
$begingroup$
@Bill Yes, you're right. I just wanted to give a different argument than the one the OP already had. I suppose I wanted to show a possible way to attack problems in which one has to prove uniqueness and there's no explicit form for the object that will show uniqueness immediately, since the OP asked for such a method. For instance when proving uniqueness of the inverse element in an arbitrary group.
$endgroup$
– Adrián Barquero
Dec 24 '10 at 20:39
$begingroup$
án: But it's still redundant even for inverses in an arbitrary group - that's merely the special case $b = 0$ above. Perhaps you meant in an arbitrary semigroup, i.e. where an inverse operation need not exist.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 21:16
add a comment |
$begingroup$
Simply use the (additive group) axioms to prove $rm a + x = b Rightarrow x = b + (-a):. $ That yields both existence and uniqueness. That the uniqueness doesn't require further proof is a subtlety that sometimes confuses students. This is discussed at length in my posts in Uniqueness of solution of x+a=b from field, sci.math, May 5, 2003 regarding a slick proof in Max Rosenlicht's Introduction to Analysis.
answered Dec 24 '10 at 19:56
Bill DubuqueBill Dubuque
213k29196654
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Usually when you want to prove something is unique you start by assuming there are two things that satisfy the given property, and then you show that they are the same. In this case, if you want to prove that there is exactly one $x$ such that $a + x = b$, you could start by assuming there are two numbers, say $x$ and $y$ that satisfy the identity, that is
$$a + x = b quad text{and} quad a + y = b$$
but then this implies that $a + x = a + y$ since both are equal to $b$. Now you can add $-a$ to both sides of the equation to get
$$-a + (a + x) = -a + (a + y) implies (-a + a) + x = (-a + a) + y $$
$$implies 0 + x = 0 + y implies x = y$$
This takes care of the uniqueness of the number and since you already know one such $x$ then you're done. I suppose you're doing this in the real number field and that you know which axioms take place in the argument. Hope this helps a little.
answered Dec 23 '10 at 23:10
Adrián BarqueroAdrián Barquero
10.8k23982
$endgroup$
$begingroup$
So proof by contradiction? But what about the given argument, is it really air tight?
$endgroup$
– rsteckly
Dec 23 '10 at 23:13
1
$begingroup$
Yes, it is correct. What I did is to assume two numbers $x, y$ satisfy the identity and proved that $x = y$. What your text does is to show what $x$ should be, so it is another way of proving uniqueness.
$endgroup$
– Adrián Barquero
Dec 23 '10 at 23:16
$begingroup$
án: But it's redundant to give such a uniqueness proof - see my answer.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 19:57
$begingroup$
@Bill Yes, you're right. I just wanted to give a different argument than the one the OP already had. I suppose I wanted to show a possible way to attack problems in which one has to prove uniqueness and there's no explicit form for the object that will show uniqueness immediately, since the OP asked for such a method. For instance when proving uniqueness of the inverse element in an arbitrary group.
$endgroup$
– Adrián Barquero
Dec 24 '10 at 20:39
$begingroup$
án: But it's still redundant even for inverses in an arbitrary group - that's merely the special case $b = 0$ above. Perhaps you meant in an arbitrary semigroup, i.e. where an inverse operation need not exist.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 21:16
add a comment |
$begingroup$
Usually when you want to prove something is unique you start by assuming there are two things that satisfy the given property, and then you show that they are the same. In this case, if you want to prove that there is exactly one $x$ such that $a + x = b$, you could start by assuming there are two numbers, say $x$ and $y$ that satisfy the identity, that is
$$a + x = b quad text{and} quad a + y = b$$
but then this implies that $a + x = a + y$ since both are equal to $b$. Now you can add $-a$ to both sides of the equation to get
$$-a + (a + x) = -a + (a + y) implies (-a + a) + x = (-a + a) + y $$
$$implies 0 + x = 0 + y implies x = y$$
This takes care of the uniqueness of the number and since you already know one such $x$ then you're done. I suppose you're doing this in the real number field and that you know which axioms take place in the argument. Hope this helps a little.
answered Dec 23 '10 at 23:10
Adrián BarqueroAdrián Barquero
10.8k23982
$endgroup$
$begingroup$
So proof by contradiction? But what about the given argument, is it really air tight?
$endgroup$
– rsteckly
Dec 23 '10 at 23:13
1
$begingroup$
Yes, it is correct. What I did is to assume two numbers $x, y$ satisfy the identity and proved that $x = y$. What your text does is to show what $x$ should be, so it is another way of proving uniqueness.
$endgroup$
– Adrián Barquero
Dec 23 '10 at 23:16
$begingroup$
án: But it's redundant to give such a uniqueness proof - see my answer.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 19:57
$begingroup$
@Bill Yes, you're right. I just wanted to give a different argument than the one the OP already had. I suppose I wanted to show a possible way to attack problems in which one has to prove uniqueness and there's no explicit form for the object that will show uniqueness immediately, since the OP asked for such a method. For instance when proving uniqueness of the inverse element in an arbitrary group.
$endgroup$
– Adrián Barquero
Dec 24 '10 at 20:39
$begingroup$
án: But it's still redundant even for inverses in an arbitrary group - that's merely the special case $b = 0$ above. Perhaps you meant in an arbitrary semigroup, i.e. where an inverse operation need not exist.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 21:16
add a comment |
$begingroup$
Usually when you want to prove something is unique you start by assuming there are two things that satisfy the given property, and then you show that they are the same. In this case, if you want to prove that there is exactly one $x$ such that $a + x = b$, you could start by assuming there are two numbers, say $x$ and $y$ that satisfy the identity, that is
$$a + x = b quad text{and} quad a + y = b$$
but then this implies that $a + x = a + y$ since both are equal to $b$. Now you can add $-a$ to both sides of the equation to get
$$-a + (a + x) = -a + (a + y) implies (-a + a) + x = (-a + a) + y $$
$$implies 0 + x = 0 + y implies x = y$$
This takes care of the uniqueness of the number and since you already know one such $x$ then you're done. I suppose you're doing this in the real number field and that you know which axioms take place in the argument. Hope this helps a little.
answered Dec 23 '10 at 23:10
Adrián BarqueroAdrián Barquero
10.8k23982
$endgroup$
Usually when you want to prove something is unique you start by assuming there are two things that satisfy the given property, and then you show that they are the same. In this case, if you want to prove that there is exactly one $x$ such that $a + x = b$, you could start by assuming there are two numbers, say $x$ and $y$ that satisfy the identity, that is
$$a + x = b quad text{and} quad a + y = b$$
but then this implies that $a + x = a + y$ since both are equal to $b$. Now you can add $-a$ to both sides of the equation to get
$$-a + (a + x) = -a + (a + y) implies (-a + a) + x = (-a + a) + y $$
$$implies 0 + x = 0 + y implies x = y$$
This takes care of the uniqueness of the number and since you already know one such $x$ then you're done. I suppose you're doing this in the real number field and that you know which axioms take place in the argument. Hope this helps a little.
answered Dec 23 '10 at 23:10
Adrián BarqueroAdrián Barquero
10.8k23982
answered Dec 23 '10 at 23:10
Adrián BarqueroAdrián Barquero
10.8k23982
answered Dec 23 '10 at 23:10
Adrián BarqueroAdrián Barquero
10.8k23982
answered Dec 23 '10 at 23:10
Adrián BarqueroAdrián Barquero
10.8k23982
10.8k23982
$begingroup$
So proof by contradiction? But what about the given argument, is it really air tight?
$endgroup$
– rsteckly
Dec 23 '10 at 23:13
1
$begingroup$
Yes, it is correct. What I did is to assume two numbers $x, y$ satisfy the identity and proved that $x = y$. What your text does is to show what $x$ should be, so it is another way of proving uniqueness.
$endgroup$
– Adrián Barquero
Dec 23 '10 at 23:16
$begingroup$
án: But it's redundant to give such a uniqueness proof - see my answer.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 19:57
$begingroup$
@Bill Yes, you're right. I just wanted to give a different argument than the one the OP already had. I suppose I wanted to show a possible way to attack problems in which one has to prove uniqueness and there's no explicit form for the object that will show uniqueness immediately, since the OP asked for such a method. For instance when proving uniqueness of the inverse element in an arbitrary group.
$endgroup$
– Adrián Barquero
Dec 24 '10 at 20:39
$begingroup$
án: But it's still redundant even for inverses in an arbitrary group - that's merely the special case $b = 0$ above. Perhaps you meant in an arbitrary semigroup, i.e. where an inverse operation need not exist.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 21:16
add a comment |
$begingroup$
So proof by contradiction? But what about the given argument, is it really air tight?
$endgroup$
– rsteckly
Dec 23 '10 at 23:13
1
$begingroup$
Yes, it is correct. What I did is to assume two numbers $x, y$ satisfy the identity and proved that $x = y$. What your text does is to show what $x$ should be, so it is another way of proving uniqueness.
$endgroup$
– Adrián Barquero
Dec 23 '10 at 23:16
$begingroup$
án: But it's redundant to give such a uniqueness proof - see my answer.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 19:57
$begingroup$
@Bill Yes, you're right. I just wanted to give a different argument than the one the OP already had. I suppose I wanted to show a possible way to attack problems in which one has to prove uniqueness and there's no explicit form for the object that will show uniqueness immediately, since the OP asked for such a method. For instance when proving uniqueness of the inverse element in an arbitrary group.
$endgroup$
– Adrián Barquero
Dec 24 '10 at 20:39
$begingroup$
án: But it's still redundant even for inverses in an arbitrary group - that's merely the special case $b = 0$ above. Perhaps you meant in an arbitrary semigroup, i.e. where an inverse operation need not exist.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 21:16
$begingroup$
So proof by contradiction? But what about the given argument, is it really air tight?
$endgroup$
– rsteckly
Dec 23 '10 at 23:13
$begingroup$
So proof by contradiction? But what about the given argument, is it really air tight?
$endgroup$
– rsteckly
Dec 23 '10 at 23:13
1
1
$begingroup$
Yes, it is correct. What I did is to assume two numbers $x, y$ satisfy the identity and proved that $x = y$. What your text does is to show what $x$ should be, so it is another way of proving uniqueness.
$endgroup$
– Adrián Barquero
Dec 23 '10 at 23:16
$begingroup$
Yes, it is correct. What I did is to assume two numbers $x, y$ satisfy the identity and proved that $x = y$. What your text does is to show what $x$ should be, so it is another way of proving uniqueness.
$endgroup$
– Adrián Barquero
Dec 23 '10 at 23:16
$begingroup$
án: But it's redundant to give such a uniqueness proof - see my answer.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 19:57
$begingroup$
án: But it's redundant to give such a uniqueness proof - see my answer.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 19:57
$begingroup$
@Bill Yes, you're right. I just wanted to give a different argument than the one the OP already had. I suppose I wanted to show a possible way to attack problems in which one has to prove uniqueness and there's no explicit form for the object that will show uniqueness immediately, since the OP asked for such a method. For instance when proving uniqueness of the inverse element in an arbitrary group.
$endgroup$
– Adrián Barquero
Dec 24 '10 at 20:39
$begingroup$
@Bill Yes, you're right. I just wanted to give a different argument than the one the OP already had. I suppose I wanted to show a possible way to attack problems in which one has to prove uniqueness and there's no explicit form for the object that will show uniqueness immediately, since the OP asked for such a method. For instance when proving uniqueness of the inverse element in an arbitrary group.
$endgroup$
– Adrián Barquero
Dec 24 '10 at 20:39
$begingroup$
án: But it's still redundant even for inverses in an arbitrary group - that's merely the special case $b = 0$ above. Perhaps you meant in an arbitrary semigroup, i.e. where an inverse operation need not exist.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 21:16
$begingroup$
án: But it's still redundant even for inverses in an arbitrary group - that's merely the special case $b = 0$ above. Perhaps you meant in an arbitrary semigroup, i.e. where an inverse operation need not exist.
$endgroup$
– Bill Dubuque
Dec 24 '10 at 21:16
add a comment |
$begingroup$
Simply use the (additive group) axioms to prove $rm a + x = b Rightarrow x = b + (-a):. $ That yields both existence and uniqueness. That the uniqueness doesn't require further proof is a subtlety that sometimes confuses students. This is discussed at length in my posts in Uniqueness of solution of x+a=b from field, sci.math, May 5, 2003 regarding a slick proof in Max Rosenlicht's Introduction to Analysis.
answered Dec 24 '10 at 19:56
Bill DubuqueBill Dubuque
213k29196654
$endgroup$
add a comment |
$begingroup$
Simply use the (additive group) axioms to prove $rm a + x = b Rightarrow x = b + (-a):. $ That yields both existence and uniqueness. That the uniqueness doesn't require further proof is a subtlety that sometimes confuses students. This is discussed at length in my posts in Uniqueness of solution of x+a=b from field, sci.math, May 5, 2003 regarding a slick proof in Max Rosenlicht's Introduction to Analysis.
answered Dec 24 '10 at 19:56
Bill DubuqueBill Dubuque
213k29196654
$endgroup$
add a comment |
$begingroup$
Simply use the (additive group) axioms to prove $rm a + x = b Rightarrow x = b + (-a):. $ That yields both existence and uniqueness. That the uniqueness doesn't require further proof is a subtlety that sometimes confuses students. This is discussed at length in my posts in Uniqueness of solution of x+a=b from field, sci.math, May 5, 2003 regarding a slick proof in Max Rosenlicht's Introduction to Analysis.
answered Dec 24 '10 at 19:56
Bill DubuqueBill Dubuque
213k29196654
$endgroup$
Simply use the (additive group) axioms to prove $rm a + x = b Rightarrow x = b + (-a):. $ That yields both existence and uniqueness. That the uniqueness doesn't require further proof is a subtlety that sometimes confuses students. This is discussed at length in my posts in Uniqueness of solution of x+a=b from field, sci.math, May 5, 2003 regarding a slick proof in Max Rosenlicht's Introduction to Analysis.
answered Dec 24 '10 at 19:56
Bill DubuqueBill Dubuque
213k29196654
edited Mar 19 at 2:37
answered Dec 24 '10 at 19:56
Bill DubuqueBill Dubuque
213k29196654
answered Dec 24 '10 at 19:56
Bill DubuqueBill Dubuque
213k29196654
answered Dec 24 '10 at 19:56
Bill DubuqueBill Dubuque
213k29196654
213k29196654
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$begingroup$
Yes, it is essentially saying that the only possible value is b+(-a).
$endgroup$
– Aryabhata
Dec 23 '10 at 23:04
$begingroup$
Relevant SMBC comic (I think)
$endgroup$
– Nick T
Jan 5 '14 at 19:21