Confusion regarding the completeness axiom
Why would we use square root of 2 in our example to show that the rationals don't have a supremum when square root of 2 is not an element of the rationals? Wouldnt the supremum be the element of the rationals that is greater square root 2 or less than square root 2?
proof-explanation
asked Nov 24 at 3:33
Imran
1004
add a comment |
Why would we use square root of 2 in our example to show that the rationals don't have a supremum when square root of 2 is not an element of the rationals? Wouldnt the supremum be the element of the rationals that is greater square root 2 or less than square root 2?
proof-explanation
asked Nov 24 at 3:33
Imran
1004
add a comment |
Why would we use square root of 2 in our example to show that the rationals don't have a supremum when square root of 2 is not an element of the rationals? Wouldnt the supremum be the element of the rationals that is greater square root 2 or less than square root 2?
proof-explanation
asked Nov 24 at 3:33
Imran
1004
Why would we use square root of 2 in our example to show that the rationals don't have a supremum when square root of 2 is not an element of the rationals? Wouldnt the supremum be the element of the rationals that is greater square root 2 or less than square root 2?
proof-explanation
proof-explanation
asked Nov 24 at 3:33
Imran
1004
asked Nov 24 at 3:33
Imran
1004
asked Nov 24 at 3:33
Imran
1004
asked Nov 24 at 3:33
Imran
1004
asked Nov 24 at 3:33
Imran
1004
1004
add a comment |
add a comment |
2 Answers
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We can still talk about the set of rationals less than $sqrt 2$ without making direct reference to $sqrt 2.$ Write it as $${qinmathbb Q: mbox{$q^2<2$ or $q<0$}}.$$
As to your second sentence, I'm not sure I understand... any element of the rationals is greater than or less than $sqrt{2}.$ Maybe you meant greater than or equal to and less than or equal to $sqrt 2$? But this just means it is equal to $sqrt{2},$ and the usual proof that $sqrt{2}$ is irrational shows that this is impossible.
Showing the set has no least upper bound in the rationals requires one more step than showing there is no rational $q$ with $q^2 =2.$ We also use the fact that the rationals are dense. We know that any upper bound must have $q^2 > 2,$ and then we can show that there is a slightly smaller rational that still squares to greater than two.
answered Nov 24 at 3:44
spaceisdarkgreen
32.3k21753
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You don't use the square root of $2$ directly, rather you look at the set of rational numbers $q$ such that $q^2 < 2$, so the definition of that set does not refer to anything other than rational numbers and multiplication of them.
answered Nov 24 at 3:47
Ned
1,948910
add a comment |
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2 Answers
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2 Answers
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We can still talk about the set of rationals less than $sqrt 2$ without making direct reference to $sqrt 2.$ Write it as $${qinmathbb Q: mbox{$q^2<2$ or $q<0$}}.$$
As to your second sentence, I'm not sure I understand... any element of the rationals is greater than or less than $sqrt{2}.$ Maybe you meant greater than or equal to and less than or equal to $sqrt 2$? But this just means it is equal to $sqrt{2},$ and the usual proof that $sqrt{2}$ is irrational shows that this is impossible.
Showing the set has no least upper bound in the rationals requires one more step than showing there is no rational $q$ with $q^2 =2.$ We also use the fact that the rationals are dense. We know that any upper bound must have $q^2 > 2,$ and then we can show that there is a slightly smaller rational that still squares to greater than two.
answered Nov 24 at 3:44
spaceisdarkgreen
32.3k21753
add a comment |
We can still talk about the set of rationals less than $sqrt 2$ without making direct reference to $sqrt 2.$ Write it as $${qinmathbb Q: mbox{$q^2<2$ or $q<0$}}.$$
As to your second sentence, I'm not sure I understand... any element of the rationals is greater than or less than $sqrt{2}.$ Maybe you meant greater than or equal to and less than or equal to $sqrt 2$? But this just means it is equal to $sqrt{2},$ and the usual proof that $sqrt{2}$ is irrational shows that this is impossible.
Showing the set has no least upper bound in the rationals requires one more step than showing there is no rational $q$ with $q^2 =2.$ We also use the fact that the rationals are dense. We know that any upper bound must have $q^2 > 2,$ and then we can show that there is a slightly smaller rational that still squares to greater than two.
answered Nov 24 at 3:44
spaceisdarkgreen
32.3k21753
add a comment |
We can still talk about the set of rationals less than $sqrt 2$ without making direct reference to $sqrt 2.$ Write it as $${qinmathbb Q: mbox{$q^2<2$ or $q<0$}}.$$
As to your second sentence, I'm not sure I understand... any element of the rationals is greater than or less than $sqrt{2}.$ Maybe you meant greater than or equal to and less than or equal to $sqrt 2$? But this just means it is equal to $sqrt{2},$ and the usual proof that $sqrt{2}$ is irrational shows that this is impossible.
Showing the set has no least upper bound in the rationals requires one more step than showing there is no rational $q$ with $q^2 =2.$ We also use the fact that the rationals are dense. We know that any upper bound must have $q^2 > 2,$ and then we can show that there is a slightly smaller rational that still squares to greater than two.
answered Nov 24 at 3:44
spaceisdarkgreen
32.3k21753
We can still talk about the set of rationals less than $sqrt 2$ without making direct reference to $sqrt 2.$ Write it as $${qinmathbb Q: mbox{$q^2<2$ or $q<0$}}.$$
As to your second sentence, I'm not sure I understand... any element of the rationals is greater than or less than $sqrt{2}.$ Maybe you meant greater than or equal to and less than or equal to $sqrt 2$? But this just means it is equal to $sqrt{2},$ and the usual proof that $sqrt{2}$ is irrational shows that this is impossible.
Showing the set has no least upper bound in the rationals requires one more step than showing there is no rational $q$ with $q^2 =2.$ We also use the fact that the rationals are dense. We know that any upper bound must have $q^2 > 2,$ and then we can show that there is a slightly smaller rational that still squares to greater than two.
answered Nov 24 at 3:44
spaceisdarkgreen
32.3k21753
edited Nov 24 at 4:30
answered Nov 24 at 3:44
spaceisdarkgreen
32.3k21753
answered Nov 24 at 3:44
spaceisdarkgreen
32.3k21753
answered Nov 24 at 3:44
spaceisdarkgreen
32.3k21753
32.3k21753
add a comment |
add a comment |
You don't use the square root of $2$ directly, rather you look at the set of rational numbers $q$ such that $q^2 < 2$, so the definition of that set does not refer to anything other than rational numbers and multiplication of them.
answered Nov 24 at 3:47
Ned
1,948910
add a comment |
You don't use the square root of $2$ directly, rather you look at the set of rational numbers $q$ such that $q^2 < 2$, so the definition of that set does not refer to anything other than rational numbers and multiplication of them.
answered Nov 24 at 3:47
Ned
1,948910
add a comment |
You don't use the square root of $2$ directly, rather you look at the set of rational numbers $q$ such that $q^2 < 2$, so the definition of that set does not refer to anything other than rational numbers and multiplication of them.
answered Nov 24 at 3:47
Ned
1,948910
You don't use the square root of $2$ directly, rather you look at the set of rational numbers $q$ such that $q^2 < 2$, so the definition of that set does not refer to anything other than rational numbers and multiplication of them.
answered Nov 24 at 3:47
Ned
1,948910
answered Nov 24 at 3:47
Ned
1,948910
answered Nov 24 at 3:47
Ned
1,948910
answered Nov 24 at 3:47
Ned
1,948910
1,948910
add a comment |
add a comment |
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