Mixing
Given $x^2=2$ prove for any rational number $frac{p}{q} < x$,there exists $frac{m}{n}$ such that...
Without using limits or the definition of irrational numbers, how do you solve this? I was thinking proof by contradiction, but I keep running into problems.
real-analysis proof-writing real-numbers rational-numbers
edited Nov 21 '18 at 4:17
Martund
1,405212
asked Nov 21 '18 at 3:49
Masie
113
add a comment |
Without using limits or the definition of irrational numbers, how do you solve this? I was thinking proof by contradiction, but I keep running into problems.
real-analysis proof-writing real-numbers rational-numbers
edited Nov 21 '18 at 4:17
Martund
1,405212
asked Nov 21 '18 at 3:49
Masie
113
add a comment |
Without using limits or the definition of irrational numbers, how do you solve this? I was thinking proof by contradiction, but I keep running into problems.
real-analysis proof-writing real-numbers rational-numbers
edited Nov 21 '18 at 4:17
Martund
1,405212
asked Nov 21 '18 at 3:49
Masie
113
Without using limits or the definition of irrational numbers, how do you solve this? I was thinking proof by contradiction, but I keep running into problems.
real-analysis proof-writing real-numbers rational-numbers
real-analysis proof-writing real-numbers rational-numbers
edited Nov 21 '18 at 4:17
Martund
1,405212
asked Nov 21 '18 at 3:49
Masie
113
edited Nov 21 '18 at 4:17
Martund
1,405212
asked Nov 21 '18 at 3:49
Masie
113
edited Nov 21 '18 at 4:17
Martund
1,405212
edited Nov 21 '18 at 4:17
Martund
1,405212
edited Nov 21 '18 at 4:17
Martund
1,405212
1,405212
asked Nov 21 '18 at 3:49
Masie
113
asked Nov 21 '18 at 3:49
Masie
113
asked Nov 21 '18 at 3:49
Masie
113
113
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
There are many possible answers, but I believe this one complies with the restrictions imposed (assuming $p,q>0$): $$frac{m}{n} = frac{4pq}{p^2+2q^2}.$$
Proof of $frac{p}{q} < frac{m}{n}$. By hypothesis $p^2 < 2q^2$ then $$frac{m}{n} = frac{4pq}{p^2+2q^2} > frac{4pq}{2q^2+2q^2} = frac{4pq}{4q^2} = frac{p}{q}.$$
Proof of $frac{m}{n} < sqrt{2}$, or equivalently, $0 < 2n^2 - m^2$:
$$2n^2 - m^2 = 2(p^2+2q^2)^2 - 16 p^4q^4 = 2 (2q^2 - p^2)^2 > 0.$$
Q.E.D.
answered Nov 21 '18 at 6:05
mlerma54
1,162138
My approach was to use Newton's method to solve the equation $x^2 = 1/2$, enter $q/p$ in the algorithm and interpret its output as $n/m$. Since $q/p > 1/sqrt{2}$ and $x^2-1/2$ is convex, then $n/m$ had to be between $1/sqrt{2}$ and $q/p$.
– mlerma54
Nov 21 '18 at 6:12
Thanks! This helped a lot!
– Masie
Nov 22 '18 at 2:30
Just out of curiosity, how did you come up with that value for m/n ?
– Masie
Nov 22 '18 at 20:35
@Masie I used Newton's method for finding approximate roots of $x^2 - 1/2$ - Wikipedia contains a detailed description of the method. If you start with $q/p$ as a first "guess", then you can take $n/m$ as the next approximation generated by the algorithm. I didn't use $x^2-2$ and $p/q$ because the next approximation would overshoot the root.
– mlerma54
Nov 23 '18 at 3:47
Ah! I see. Got it. Thanks
– Masie
Nov 27 '18 at 20:37
add a comment |
Let $a$ be a rational, close to, but below $sqrt2$. Then $b=2/a$ is a rational,
close to, but above $sqrt2$. Consider $c=frac12(a+b)$. That is rational
and should be even closer to $sqrt2$. But it turns out that $c>sqrt2$. Why
not try then $d=2/c$? Can you prove $a<d<sqrt2$?
answered Nov 21 '18 at 4:05
Lord Shark the Unknown
101k958132
add a comment |
We don't need the definition of irrational numbers, but we know that x is real. Now, by density of rational numbers, we can say that for any real number not equal to x ($frac{p}{q}$ in particular) their exist a rational between the number and x. Call it $frac{m}{n}$ and we are done.
Hope it is helpful.
answered Nov 21 '18 at 4:05
Martund
1,405212
I sincerely doubt that the density of rational numbers nor the existence of a real $sqrt 2$ is expected to be known for this exercise.
– fleablood
Nov 21 '18 at 4:25
add a comment |
Let $frac{p}{q}=r$ and
if $r>sqrt{2} - 1$, consider $(r+h)^2<2$ where $0<h<1$.
Then $r^2+2rh+h^2<2.$ Let $r^2+2hr+h=2 <r^2+2rh+h^2$.
Then $h=frac{2-r^2}{2r+1}<1$
if $r<sqrt{2} - 1$, take $h=1$.
answered Nov 21 '18 at 4:28
Offlaw
2649
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
There are many possible answers, but I believe this one complies with the restrictions imposed (assuming $p,q>0$): $$frac{m}{n} = frac{4pq}{p^2+2q^2}.$$
Proof of $frac{p}{q} < frac{m}{n}$. By hypothesis $p^2 < 2q^2$ then $$frac{m}{n} = frac{4pq}{p^2+2q^2} > frac{4pq}{2q^2+2q^2} = frac{4pq}{4q^2} = frac{p}{q}.$$
Proof of $frac{m}{n} < sqrt{2}$, or equivalently, $0 < 2n^2 - m^2$:
$$2n^2 - m^2 = 2(p^2+2q^2)^2 - 16 p^4q^4 = 2 (2q^2 - p^2)^2 > 0.$$
Q.E.D.
answered Nov 21 '18 at 6:05
mlerma54
1,162138
My approach was to use Newton's method to solve the equation $x^2 = 1/2$, enter $q/p$ in the algorithm and interpret its output as $n/m$. Since $q/p > 1/sqrt{2}$ and $x^2-1/2$ is convex, then $n/m$ had to be between $1/sqrt{2}$ and $q/p$.
– mlerma54
Nov 21 '18 at 6:12
Thanks! This helped a lot!
– Masie
Nov 22 '18 at 2:30
Just out of curiosity, how did you come up with that value for m/n ?
– Masie
Nov 22 '18 at 20:35
@Masie I used Newton's method for finding approximate roots of $x^2 - 1/2$ - Wikipedia contains a detailed description of the method. If you start with $q/p$ as a first "guess", then you can take $n/m$ as the next approximation generated by the algorithm. I didn't use $x^2-2$ and $p/q$ because the next approximation would overshoot the root.
– mlerma54
Nov 23 '18 at 3:47
Ah! I see. Got it. Thanks
– Masie
Nov 27 '18 at 20:37
add a comment |
There are many possible answers, but I believe this one complies with the restrictions imposed (assuming $p,q>0$): $$frac{m}{n} = frac{4pq}{p^2+2q^2}.$$
Proof of $frac{p}{q} < frac{m}{n}$. By hypothesis $p^2 < 2q^2$ then $$frac{m}{n} = frac{4pq}{p^2+2q^2} > frac{4pq}{2q^2+2q^2} = frac{4pq}{4q^2} = frac{p}{q}.$$
Proof of $frac{m}{n} < sqrt{2}$, or equivalently, $0 < 2n^2 - m^2$:
$$2n^2 - m^2 = 2(p^2+2q^2)^2 - 16 p^4q^4 = 2 (2q^2 - p^2)^2 > 0.$$
Q.E.D.
answered Nov 21 '18 at 6:05
mlerma54
1,162138
My approach was to use Newton's method to solve the equation $x^2 = 1/2$, enter $q/p$ in the algorithm and interpret its output as $n/m$. Since $q/p > 1/sqrt{2}$ and $x^2-1/2$ is convex, then $n/m$ had to be between $1/sqrt{2}$ and $q/p$.
– mlerma54
Nov 21 '18 at 6:12
Thanks! This helped a lot!
– Masie
Nov 22 '18 at 2:30
Just out of curiosity, how did you come up with that value for m/n ?
– Masie
Nov 22 '18 at 20:35
@Masie I used Newton's method for finding approximate roots of $x^2 - 1/2$ - Wikipedia contains a detailed description of the method. If you start with $q/p$ as a first "guess", then you can take $n/m$ as the next approximation generated by the algorithm. I didn't use $x^2-2$ and $p/q$ because the next approximation would overshoot the root.
– mlerma54
Nov 23 '18 at 3:47
Ah! I see. Got it. Thanks
– Masie
Nov 27 '18 at 20:37
add a comment |
There are many possible answers, but I believe this one complies with the restrictions imposed (assuming $p,q>0$): $$frac{m}{n} = frac{4pq}{p^2+2q^2}.$$
Proof of $frac{p}{q} < frac{m}{n}$. By hypothesis $p^2 < 2q^2$ then $$frac{m}{n} = frac{4pq}{p^2+2q^2} > frac{4pq}{2q^2+2q^2} = frac{4pq}{4q^2} = frac{p}{q}.$$
Proof of $frac{m}{n} < sqrt{2}$, or equivalently, $0 < 2n^2 - m^2$:
$$2n^2 - m^2 = 2(p^2+2q^2)^2 - 16 p^4q^4 = 2 (2q^2 - p^2)^2 > 0.$$
Q.E.D.
answered Nov 21 '18 at 6:05
mlerma54
1,162138
There are many possible answers, but I believe this one complies with the restrictions imposed (assuming $p,q>0$): $$frac{m}{n} = frac{4pq}{p^2+2q^2}.$$
Proof of $frac{p}{q} < frac{m}{n}$. By hypothesis $p^2 < 2q^2$ then $$frac{m}{n} = frac{4pq}{p^2+2q^2} > frac{4pq}{2q^2+2q^2} = frac{4pq}{4q^2} = frac{p}{q}.$$
Proof of $frac{m}{n} < sqrt{2}$, or equivalently, $0 < 2n^2 - m^2$:
$$2n^2 - m^2 = 2(p^2+2q^2)^2 - 16 p^4q^4 = 2 (2q^2 - p^2)^2 > 0.$$
Q.E.D.
answered Nov 21 '18 at 6:05
mlerma54
1,162138
answered Nov 21 '18 at 6:05
mlerma54
1,162138
answered Nov 21 '18 at 6:05
mlerma54
1,162138
answered Nov 21 '18 at 6:05
mlerma54
1,162138
1,162138
My approach was to use Newton's method to solve the equation $x^2 = 1/2$, enter $q/p$ in the algorithm and interpret its output as $n/m$. Since $q/p > 1/sqrt{2}$ and $x^2-1/2$ is convex, then $n/m$ had to be between $1/sqrt{2}$ and $q/p$.
– mlerma54
Nov 21 '18 at 6:12
Thanks! This helped a lot!
– Masie
Nov 22 '18 at 2:30
Just out of curiosity, how did you come up with that value for m/n ?
– Masie
Nov 22 '18 at 20:35
@Masie I used Newton's method for finding approximate roots of $x^2 - 1/2$ - Wikipedia contains a detailed description of the method. If you start with $q/p$ as a first "guess", then you can take $n/m$ as the next approximation generated by the algorithm. I didn't use $x^2-2$ and $p/q$ because the next approximation would overshoot the root.
– mlerma54
Nov 23 '18 at 3:47
Ah! I see. Got it. Thanks
– Masie
Nov 27 '18 at 20:37
add a comment |
My approach was to use Newton's method to solve the equation $x^2 = 1/2$, enter $q/p$ in the algorithm and interpret its output as $n/m$. Since $q/p > 1/sqrt{2}$ and $x^2-1/2$ is convex, then $n/m$ had to be between $1/sqrt{2}$ and $q/p$.
– mlerma54
Nov 21 '18 at 6:12
Thanks! This helped a lot!
– Masie
Nov 22 '18 at 2:30
Just out of curiosity, how did you come up with that value for m/n ?
– Masie
Nov 22 '18 at 20:35
@Masie I used Newton's method for finding approximate roots of $x^2 - 1/2$ - Wikipedia contains a detailed description of the method. If you start with $q/p$ as a first "guess", then you can take $n/m$ as the next approximation generated by the algorithm. I didn't use $x^2-2$ and $p/q$ because the next approximation would overshoot the root.
– mlerma54
Nov 23 '18 at 3:47
Ah! I see. Got it. Thanks
– Masie
Nov 27 '18 at 20:37
My approach was to use Newton's method to solve the equation $x^2 = 1/2$, enter $q/p$ in the algorithm and interpret its output as $n/m$. Since $q/p > 1/sqrt{2}$ and $x^2-1/2$ is convex, then $n/m$ had to be between $1/sqrt{2}$ and $q/p$.
– mlerma54
Nov 21 '18 at 6:12
My approach was to use Newton's method to solve the equation $x^2 = 1/2$, enter $q/p$ in the algorithm and interpret its output as $n/m$. Since $q/p > 1/sqrt{2}$ and $x^2-1/2$ is convex, then $n/m$ had to be between $1/sqrt{2}$ and $q/p$.
– mlerma54
Nov 21 '18 at 6:12
Thanks! This helped a lot!
– Masie
Nov 22 '18 at 2:30
Thanks! This helped a lot!
– Masie
Nov 22 '18 at 2:30
Just out of curiosity, how did you come up with that value for m/n ?
– Masie
Nov 22 '18 at 20:35
Just out of curiosity, how did you come up with that value for m/n ?
– Masie
Nov 22 '18 at 20:35
@Masie I used Newton's method for finding approximate roots of $x^2 - 1/2$ - Wikipedia contains a detailed description of the method. If you start with $q/p$ as a first "guess", then you can take $n/m$ as the next approximation generated by the algorithm. I didn't use $x^2-2$ and $p/q$ because the next approximation would overshoot the root.
– mlerma54
Nov 23 '18 at 3:47
@Masie I used Newton's method for finding approximate roots of $x^2 - 1/2$ - Wikipedia contains a detailed description of the method. If you start with $q/p$ as a first "guess", then you can take $n/m$ as the next approximation generated by the algorithm. I didn't use $x^2-2$ and $p/q$ because the next approximation would overshoot the root.
– mlerma54
Nov 23 '18 at 3:47
Ah! I see. Got it. Thanks
– Masie
Nov 27 '18 at 20:37
Ah! I see. Got it. Thanks
– Masie
Nov 27 '18 at 20:37
add a comment |
Let $a$ be a rational, close to, but below $sqrt2$. Then $b=2/a$ is a rational,
close to, but above $sqrt2$. Consider $c=frac12(a+b)$. That is rational
and should be even closer to $sqrt2$. But it turns out that $c>sqrt2$. Why
not try then $d=2/c$? Can you prove $a<d<sqrt2$?
answered Nov 21 '18 at 4:05
Lord Shark the Unknown
101k958132
add a comment |
Let $a$ be a rational, close to, but below $sqrt2$. Then $b=2/a$ is a rational,
close to, but above $sqrt2$. Consider $c=frac12(a+b)$. That is rational
and should be even closer to $sqrt2$. But it turns out that $c>sqrt2$. Why
not try then $d=2/c$? Can you prove $a<d<sqrt2$?
answered Nov 21 '18 at 4:05
Lord Shark the Unknown
101k958132
add a comment |
Let $a$ be a rational, close to, but below $sqrt2$. Then $b=2/a$ is a rational,
close to, but above $sqrt2$. Consider $c=frac12(a+b)$. That is rational
and should be even closer to $sqrt2$. But it turns out that $c>sqrt2$. Why
not try then $d=2/c$? Can you prove $a<d<sqrt2$?
answered Nov 21 '18 at 4:05
Lord Shark the Unknown
101k958132
Let $a$ be a rational, close to, but below $sqrt2$. Then $b=2/a$ is a rational,
close to, but above $sqrt2$. Consider $c=frac12(a+b)$. That is rational
and should be even closer to $sqrt2$. But it turns out that $c>sqrt2$. Why
not try then $d=2/c$? Can you prove $a<d<sqrt2$?
answered Nov 21 '18 at 4:05
Lord Shark the Unknown
101k958132
answered Nov 21 '18 at 4:05
Lord Shark the Unknown
101k958132
answered Nov 21 '18 at 4:05
Lord Shark the Unknown
101k958132
answered Nov 21 '18 at 4:05
Lord Shark the Unknown
101k958132
101k958132
add a comment |
add a comment |
We don't need the definition of irrational numbers, but we know that x is real. Now, by density of rational numbers, we can say that for any real number not equal to x ($frac{p}{q}$ in particular) their exist a rational between the number and x. Call it $frac{m}{n}$ and we are done.
Hope it is helpful.
answered Nov 21 '18 at 4:05
Martund
1,405212
I sincerely doubt that the density of rational numbers nor the existence of a real $sqrt 2$ is expected to be known for this exercise.
– fleablood
Nov 21 '18 at 4:25
add a comment |
We don't need the definition of irrational numbers, but we know that x is real. Now, by density of rational numbers, we can say that for any real number not equal to x ($frac{p}{q}$ in particular) their exist a rational between the number and x. Call it $frac{m}{n}$ and we are done.
Hope it is helpful.
answered Nov 21 '18 at 4:05
Martund
1,405212
I sincerely doubt that the density of rational numbers nor the existence of a real $sqrt 2$ is expected to be known for this exercise.
– fleablood
Nov 21 '18 at 4:25
add a comment |
We don't need the definition of irrational numbers, but we know that x is real. Now, by density of rational numbers, we can say that for any real number not equal to x ($frac{p}{q}$ in particular) their exist a rational between the number and x. Call it $frac{m}{n}$ and we are done.
Hope it is helpful.
answered Nov 21 '18 at 4:05
Martund
1,405212
We don't need the definition of irrational numbers, but we know that x is real. Now, by density of rational numbers, we can say that for any real number not equal to x ($frac{p}{q}$ in particular) their exist a rational between the number and x. Call it $frac{m}{n}$ and we are done.
Hope it is helpful.
answered Nov 21 '18 at 4:05
Martund
1,405212
answered Nov 21 '18 at 4:05
Martund
1,405212
answered Nov 21 '18 at 4:05
Martund
1,405212
answered Nov 21 '18 at 4:05
Martund
1,405212
1,405212
I sincerely doubt that the density of rational numbers nor the existence of a real $sqrt 2$ is expected to be known for this exercise.
– fleablood
Nov 21 '18 at 4:25
add a comment |
I sincerely doubt that the density of rational numbers nor the existence of a real $sqrt 2$ is expected to be known for this exercise.
– fleablood
Nov 21 '18 at 4:25
I sincerely doubt that the density of rational numbers nor the existence of a real $sqrt 2$ is expected to be known for this exercise.
– fleablood
Nov 21 '18 at 4:25
I sincerely doubt that the density of rational numbers nor the existence of a real $sqrt 2$ is expected to be known for this exercise.
– fleablood
Nov 21 '18 at 4:25
add a comment |
Let $frac{p}{q}=r$ and
if $r>sqrt{2} - 1$, consider $(r+h)^2<2$ where $0<h<1$.
Then $r^2+2rh+h^2<2.$ Let $r^2+2hr+h=2 <r^2+2rh+h^2$.
Then $h=frac{2-r^2}{2r+1}<1$
if $r<sqrt{2} - 1$, take $h=1$.
answered Nov 21 '18 at 4:28
Offlaw
2649
add a comment |
Let $frac{p}{q}=r$ and
if $r>sqrt{2} - 1$, consider $(r+h)^2<2$ where $0<h<1$.
Then $r^2+2rh+h^2<2.$ Let $r^2+2hr+h=2 <r^2+2rh+h^2$.
Then $h=frac{2-r^2}{2r+1}<1$
if $r<sqrt{2} - 1$, take $h=1$.
answered Nov 21 '18 at 4:28
Offlaw
2649
add a comment |
Let $frac{p}{q}=r$ and
if $r>sqrt{2} - 1$, consider $(r+h)^2<2$ where $0<h<1$.
Then $r^2+2rh+h^2<2.$ Let $r^2+2hr+h=2 <r^2+2rh+h^2$.
Then $h=frac{2-r^2}{2r+1}<1$
if $r<sqrt{2} - 1$, take $h=1$.
answered Nov 21 '18 at 4:28
Offlaw
2649
Let $frac{p}{q}=r$ and
if $r>sqrt{2} - 1$, consider $(r+h)^2<2$ where $0<h<1$.
Then $r^2+2rh+h^2<2.$ Let $r^2+2hr+h=2 <r^2+2rh+h^2$.
Then $h=frac{2-r^2}{2r+1}<1$
if $r<sqrt{2} - 1$, take $h=1$.
answered Nov 21 '18 at 4:28
Offlaw
2649
answered Nov 21 '18 at 4:28
Offlaw
2649
answered Nov 21 '18 at 4:28
Offlaw
2649
answered Nov 21 '18 at 4:28
Offlaw
2649
2649
add a comment |
add a comment |
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