Mixing
Irrationality of $(a_1+sqrt{b_1})(a_2+sqrt{b_2})$
$begingroup$
Sorry, for a rather silly question.
Suppose $a_1$, $b_1$, $a_2$, $b_2$ are integers, all different from zero, while $b_1$ and $b_2$ are co-prime positive integers, neither being a complete square.
Is there an elementary proof that $(a_1+sqrt{b_1})(a_2+sqrt{b_2})$ is irrational? Or maybe it's just wrong?
Multiplying gives three distinct surds $n_1sqrt{b_1}$, $n_2sqrt{b_2}$, $n_3sqrt{b_1;b_2}$, so it doesn't seem to help, while taking a power gives again a product of two irrationals.
How about more general case $(a_1+sqrt[m_1]{b_1})(a_2+sqrt[m_2]{b_2})$ ?
roots irrational-numbers
asked Jan 15 at 22:40
cyanidecyanide
429217
$endgroup$
add a comment |
$begingroup$
Sorry, for a rather silly question.
Suppose $a_1$, $b_1$, $a_2$, $b_2$ are integers, all different from zero, while $b_1$ and $b_2$ are co-prime positive integers, neither being a complete square.
Is there an elementary proof that $(a_1+sqrt{b_1})(a_2+sqrt{b_2})$ is irrational? Or maybe it's just wrong?
Multiplying gives three distinct surds $n_1sqrt{b_1}$, $n_2sqrt{b_2}$, $n_3sqrt{b_1;b_2}$, so it doesn't seem to help, while taking a power gives again a product of two irrationals.
How about more general case $(a_1+sqrt[m_1]{b_1})(a_2+sqrt[m_2]{b_2})$ ?
roots irrational-numbers
asked Jan 15 at 22:40
cyanidecyanide
429217
$endgroup$
add a comment |
$begingroup$
Sorry, for a rather silly question.
Suppose $a_1$, $b_1$, $a_2$, $b_2$ are integers, all different from zero, while $b_1$ and $b_2$ are co-prime positive integers, neither being a complete square.
Is there an elementary proof that $(a_1+sqrt{b_1})(a_2+sqrt{b_2})$ is irrational? Or maybe it's just wrong?
Multiplying gives three distinct surds $n_1sqrt{b_1}$, $n_2sqrt{b_2}$, $n_3sqrt{b_1;b_2}$, so it doesn't seem to help, while taking a power gives again a product of two irrationals.
How about more general case $(a_1+sqrt[m_1]{b_1})(a_2+sqrt[m_2]{b_2})$ ?
roots irrational-numbers
asked Jan 15 at 22:40
cyanidecyanide
429217
$endgroup$
Sorry, for a rather silly question.
Suppose $a_1$, $b_1$, $a_2$, $b_2$ are integers, all different from zero, while $b_1$ and $b_2$ are co-prime positive integers, neither being a complete square.
Is there an elementary proof that $(a_1+sqrt{b_1})(a_2+sqrt{b_2})$ is irrational? Or maybe it's just wrong?
Multiplying gives three distinct surds $n_1sqrt{b_1}$, $n_2sqrt{b_2}$, $n_3sqrt{b_1;b_2}$, so it doesn't seem to help, while taking a power gives again a product of two irrationals.
How about more general case $(a_1+sqrt[m_1]{b_1})(a_2+sqrt[m_2]{b_2})$ ?
roots irrational-numbers
roots irrational-numbers
asked Jan 15 at 22:40
cyanidecyanide
429217
asked Jan 15 at 22:40
cyanidecyanide
429217
asked Jan 15 at 22:40
cyanidecyanide
429217
asked Jan 15 at 22:40
cyanidecyanide
429217
asked Jan 15 at 22:40
cyanidecyanide
429217
429217
add a comment |
add a comment |
3 Answers
3
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oldest
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$begingroup$
Suppose $(a+sqrt{b})(c+sqrt{d})=ac+asqrt d+csqrt b+sqrt{bd}=pinBbb Q.$
Then begin{align} da^2+bc^2+bd+2(acsqrt{bd}+adsqrt b+bcsqrt d) &=(p-ac)^2 \ acsqrt{bd}+adsqrt b+bcsqrt d &=q:=frac{(p-ac)^2-da^2-bc^2-bd}{2} \ sqrt{bd}(ac+asqrt d+csqrt b)=sqrt{bd}(p-sqrt{bd}) &=q \ sqrt{bd}=frac{q+bd}{p}in Bbb Qend{align}which is impossible given that $b$ and $d$ are coprime.
answered Jan 15 at 23:38
LearnerLearner
17510
$endgroup$
$begingroup$
That's the elementary proof, I was looking for. Unfortunately, it cannot be generalized for more roots, or for a higher degree.
$endgroup$
– cyanide
Jan 17 at 1:07
$begingroup$
@cyanide The argument in my Lemma is actually simpler (and it does generalize to more roots - see the link there). I updated my answer to eliminate all use of field theory to present it in elementary form.
$endgroup$
– Bill Dubuque
Jan 17 at 3:14
add a comment |
$begingroup$
Notice $ sqrt a = r + ssqrt b $ follows from $ c + sqrt{a} = dfrac{e}{d+sqrt b},$ by rationalizing the denominator.
Squaring $, a = r^2!+s^2b + 2rs, sqrt b $ so $ rs = 0,,$ else solving for $sqrt b,$ shows $,sqrt bin Bbb Q$
$sneq 0,$ else $,sqrt a = rin Bbb Q,,$ so $,r=0,$ so $,sqrt a = ssqrt b,$ $overset{largetimessqrt b}Longrightarrowsqrt ab = sbin Bbb Q,,$ contra $a,b$ coprime.
Remark $ $ Above is a special case of the Lemma below, which generalizes to any number of sqrts). See the citations there for generalizations to $n$'th roots.
Lemma $rm [K(sqrt{a},sqrt{b}) : K] = 4 $ if $rm sqrt{a}, sqrt{b}, sqrt{a,b} $ all are not in $rm,K,$ and $rm, 2 ne 0,$ in the field $rm,K.$
Proof $ $ Let $rm L = K(sqrt{b}).,$ Then $rm, [L:K] = 2,$ via $rm,sqrt{b} notin K,,$ so it suffices to prove $rm, [L(sqrt{a}):L] = 2.,$ It fails only if $rm,sqrt{a} in L = K(sqrt{b}), $ and then $rm, sqrt{a} = r + s, sqrt{b} $ for $rm r,sin K.,$ But that's impossible,
since squaring $Rightarrow rm(1)!: a = r^2 + b s^2 + 2,r,s sqrt{b},, $ which contradicts hypotheses as follows:
$rmqquadqquad rs ne 0 Rightarrow sqrt{b} in K $ by solving $(1)$ for $rmsqrt{b},,,$ using $rm,2 ne 0$
$rmqquadqquad s = 0 Rightarrow sqrt{a} in K $ via $rm sqrt{a} = r+s,sqrt b = r in K$
$rmqquadqquad r = 0 Rightarrow sqrt{a,b}in K $ via $rm sqrt{a} = s, sqrt{b}, $times $rm,sqrt{b}quad$
answered Jan 15 at 23:27
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
$begingroup$
Now, it's much clearer. This way it looks even simpler, than the one marked as an answer. Keep +1 !
$endgroup$
– cyanide
Jan 17 at 4:00
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@cyanide By clicking on the green tick you may unaccept the currently accepted answer, and then accept this one, if you've come to prefer it
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– Learner
Jan 19 at 8:19
add a comment |
$begingroup$
$(a + sqrt b)(c + sqrt d) = k in mathbb Q$ would mean
$sqrt{b} = frac k{c+sqrt d} - a$
$b = ( frac k{c+sqrt d} - a)^2 in mathbb Q$ which can probably be proven false.
Indeed $( frac k{c+sqrt d} - a)^2 = $
$frac {k(c - sqrt d)}{c^2 - d} -a)^2 =$
$(msqrt d - n)^2$ where $m = frac k{c^2-d}in mathbb Q$ and $n =frac {kc}{c^2 -d} -a in mathbb Q$.
And $(msqrt d -n)^2= m^2d -n^2 - 2nmsqrt d$ which is not rational.
Unless $n$ or $m$ is $0$.
As $kne 0$, $m ne 0$. $n = 0$ if $a = frac {kc}{c^2-d}$.
Hmmm....
answered Jan 15 at 22:51
fleabloodfleablood
71.2k22686
$endgroup$
3
$begingroup$
As for your last comment: e.g. $(-1+sqrt2)(1+sqrt2)=1$.
$endgroup$
– Learner
Jan 15 at 23:10
add a comment |
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3 Answers
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3 Answers
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oldest
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$begingroup$
Suppose $(a+sqrt{b})(c+sqrt{d})=ac+asqrt d+csqrt b+sqrt{bd}=pinBbb Q.$
Then begin{align} da^2+bc^2+bd+2(acsqrt{bd}+adsqrt b+bcsqrt d) &=(p-ac)^2 \ acsqrt{bd}+adsqrt b+bcsqrt d &=q:=frac{(p-ac)^2-da^2-bc^2-bd}{2} \ sqrt{bd}(ac+asqrt d+csqrt b)=sqrt{bd}(p-sqrt{bd}) &=q \ sqrt{bd}=frac{q+bd}{p}in Bbb Qend{align}which is impossible given that $b$ and $d$ are coprime.
answered Jan 15 at 23:38
LearnerLearner
17510
$endgroup$
$begingroup$
That's the elementary proof, I was looking for. Unfortunately, it cannot be generalized for more roots, or for a higher degree.
$endgroup$
– cyanide
Jan 17 at 1:07
$begingroup$
@cyanide The argument in my Lemma is actually simpler (and it does generalize to more roots - see the link there). I updated my answer to eliminate all use of field theory to present it in elementary form.
$endgroup$
– Bill Dubuque
Jan 17 at 3:14
add a comment |
$begingroup$
Suppose $(a+sqrt{b})(c+sqrt{d})=ac+asqrt d+csqrt b+sqrt{bd}=pinBbb Q.$
Then begin{align} da^2+bc^2+bd+2(acsqrt{bd}+adsqrt b+bcsqrt d) &=(p-ac)^2 \ acsqrt{bd}+adsqrt b+bcsqrt d &=q:=frac{(p-ac)^2-da^2-bc^2-bd}{2} \ sqrt{bd}(ac+asqrt d+csqrt b)=sqrt{bd}(p-sqrt{bd}) &=q \ sqrt{bd}=frac{q+bd}{p}in Bbb Qend{align}which is impossible given that $b$ and $d$ are coprime.
answered Jan 15 at 23:38
LearnerLearner
17510
$endgroup$
$begingroup$
That's the elementary proof, I was looking for. Unfortunately, it cannot be generalized for more roots, or for a higher degree.
$endgroup$
– cyanide
Jan 17 at 1:07
$begingroup$
@cyanide The argument in my Lemma is actually simpler (and it does generalize to more roots - see the link there). I updated my answer to eliminate all use of field theory to present it in elementary form.
$endgroup$
– Bill Dubuque
Jan 17 at 3:14
add a comment |
$begingroup$
Suppose $(a+sqrt{b})(c+sqrt{d})=ac+asqrt d+csqrt b+sqrt{bd}=pinBbb Q.$
Then begin{align} da^2+bc^2+bd+2(acsqrt{bd}+adsqrt b+bcsqrt d) &=(p-ac)^2 \ acsqrt{bd}+adsqrt b+bcsqrt d &=q:=frac{(p-ac)^2-da^2-bc^2-bd}{2} \ sqrt{bd}(ac+asqrt d+csqrt b)=sqrt{bd}(p-sqrt{bd}) &=q \ sqrt{bd}=frac{q+bd}{p}in Bbb Qend{align}which is impossible given that $b$ and $d$ are coprime.
answered Jan 15 at 23:38
LearnerLearner
17510
$endgroup$
Suppose $(a+sqrt{b})(c+sqrt{d})=ac+asqrt d+csqrt b+sqrt{bd}=pinBbb Q.$
Then begin{align} da^2+bc^2+bd+2(acsqrt{bd}+adsqrt b+bcsqrt d) &=(p-ac)^2 \ acsqrt{bd}+adsqrt b+bcsqrt d &=q:=frac{(p-ac)^2-da^2-bc^2-bd}{2} \ sqrt{bd}(ac+asqrt d+csqrt b)=sqrt{bd}(p-sqrt{bd}) &=q \ sqrt{bd}=frac{q+bd}{p}in Bbb Qend{align}which is impossible given that $b$ and $d$ are coprime.
answered Jan 15 at 23:38
LearnerLearner
17510
answered Jan 15 at 23:38
LearnerLearner
17510
answered Jan 15 at 23:38
LearnerLearner
17510
answered Jan 15 at 23:38
LearnerLearner
17510
17510
$begingroup$
That's the elementary proof, I was looking for. Unfortunately, it cannot be generalized for more roots, or for a higher degree.
$endgroup$
– cyanide
Jan 17 at 1:07
$begingroup$
@cyanide The argument in my Lemma is actually simpler (and it does generalize to more roots - see the link there). I updated my answer to eliminate all use of field theory to present it in elementary form.
$endgroup$
– Bill Dubuque
Jan 17 at 3:14
add a comment |
$begingroup$
That's the elementary proof, I was looking for. Unfortunately, it cannot be generalized for more roots, or for a higher degree.
$endgroup$
– cyanide
Jan 17 at 1:07
$begingroup$
@cyanide The argument in my Lemma is actually simpler (and it does generalize to more roots - see the link there). I updated my answer to eliminate all use of field theory to present it in elementary form.
$endgroup$
– Bill Dubuque
Jan 17 at 3:14
$begingroup$
That's the elementary proof, I was looking for. Unfortunately, it cannot be generalized for more roots, or for a higher degree.
$endgroup$
– cyanide
Jan 17 at 1:07
$begingroup$
That's the elementary proof, I was looking for. Unfortunately, it cannot be generalized for more roots, or for a higher degree.
$endgroup$
– cyanide
Jan 17 at 1:07
$begingroup$
@cyanide The argument in my Lemma is actually simpler (and it does generalize to more roots - see the link there). I updated my answer to eliminate all use of field theory to present it in elementary form.
$endgroup$
– Bill Dubuque
Jan 17 at 3:14
$begingroup$
@cyanide The argument in my Lemma is actually simpler (and it does generalize to more roots - see the link there). I updated my answer to eliminate all use of field theory to present it in elementary form.
$endgroup$
– Bill Dubuque
Jan 17 at 3:14
add a comment |
$begingroup$
Notice $ sqrt a = r + ssqrt b $ follows from $ c + sqrt{a} = dfrac{e}{d+sqrt b},$ by rationalizing the denominator.
Squaring $, a = r^2!+s^2b + 2rs, sqrt b $ so $ rs = 0,,$ else solving for $sqrt b,$ shows $,sqrt bin Bbb Q$
$sneq 0,$ else $,sqrt a = rin Bbb Q,,$ so $,r=0,$ so $,sqrt a = ssqrt b,$ $overset{largetimessqrt b}Longrightarrowsqrt ab = sbin Bbb Q,,$ contra $a,b$ coprime.
Remark $ $ Above is a special case of the Lemma below, which generalizes to any number of sqrts). See the citations there for generalizations to $n$'th roots.
Lemma $rm [K(sqrt{a},sqrt{b}) : K] = 4 $ if $rm sqrt{a}, sqrt{b}, sqrt{a,b} $ all are not in $rm,K,$ and $rm, 2 ne 0,$ in the field $rm,K.$
Proof $ $ Let $rm L = K(sqrt{b}).,$ Then $rm, [L:K] = 2,$ via $rm,sqrt{b} notin K,,$ so it suffices to prove $rm, [L(sqrt{a}):L] = 2.,$ It fails only if $rm,sqrt{a} in L = K(sqrt{b}), $ and then $rm, sqrt{a} = r + s, sqrt{b} $ for $rm r,sin K.,$ But that's impossible,
since squaring $Rightarrow rm(1)!: a = r^2 + b s^2 + 2,r,s sqrt{b},, $ which contradicts hypotheses as follows:
$rmqquadqquad rs ne 0 Rightarrow sqrt{b} in K $ by solving $(1)$ for $rmsqrt{b},,,$ using $rm,2 ne 0$
$rmqquadqquad s = 0 Rightarrow sqrt{a} in K $ via $rm sqrt{a} = r+s,sqrt b = r in K$
$rmqquadqquad r = 0 Rightarrow sqrt{a,b}in K $ via $rm sqrt{a} = s, sqrt{b}, $times $rm,sqrt{b}quad$
answered Jan 15 at 23:27
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
$begingroup$
Now, it's much clearer. This way it looks even simpler, than the one marked as an answer. Keep +1 !
$endgroup$
– cyanide
Jan 17 at 4:00
$begingroup$
@cyanide By clicking on the green tick you may unaccept the currently accepted answer, and then accept this one, if you've come to prefer it
$endgroup$
– Learner
Jan 19 at 8:19
add a comment |
$begingroup$
Notice $ sqrt a = r + ssqrt b $ follows from $ c + sqrt{a} = dfrac{e}{d+sqrt b},$ by rationalizing the denominator.
Squaring $, a = r^2!+s^2b + 2rs, sqrt b $ so $ rs = 0,,$ else solving for $sqrt b,$ shows $,sqrt bin Bbb Q$
$sneq 0,$ else $,sqrt a = rin Bbb Q,,$ so $,r=0,$ so $,sqrt a = ssqrt b,$ $overset{largetimessqrt b}Longrightarrowsqrt ab = sbin Bbb Q,,$ contra $a,b$ coprime.
Remark $ $ Above is a special case of the Lemma below, which generalizes to any number of sqrts). See the citations there for generalizations to $n$'th roots.
Lemma $rm [K(sqrt{a},sqrt{b}) : K] = 4 $ if $rm sqrt{a}, sqrt{b}, sqrt{a,b} $ all are not in $rm,K,$ and $rm, 2 ne 0,$ in the field $rm,K.$
Proof $ $ Let $rm L = K(sqrt{b}).,$ Then $rm, [L:K] = 2,$ via $rm,sqrt{b} notin K,,$ so it suffices to prove $rm, [L(sqrt{a}):L] = 2.,$ It fails only if $rm,sqrt{a} in L = K(sqrt{b}), $ and then $rm, sqrt{a} = r + s, sqrt{b} $ for $rm r,sin K.,$ But that's impossible,
since squaring $Rightarrow rm(1)!: a = r^2 + b s^2 + 2,r,s sqrt{b},, $ which contradicts hypotheses as follows:
$rmqquadqquad rs ne 0 Rightarrow sqrt{b} in K $ by solving $(1)$ for $rmsqrt{b},,,$ using $rm,2 ne 0$
$rmqquadqquad s = 0 Rightarrow sqrt{a} in K $ via $rm sqrt{a} = r+s,sqrt b = r in K$
$rmqquadqquad r = 0 Rightarrow sqrt{a,b}in K $ via $rm sqrt{a} = s, sqrt{b}, $times $rm,sqrt{b}quad$
answered Jan 15 at 23:27
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
$begingroup$
Now, it's much clearer. This way it looks even simpler, than the one marked as an answer. Keep +1 !
$endgroup$
– cyanide
Jan 17 at 4:00
$begingroup$
@cyanide By clicking on the green tick you may unaccept the currently accepted answer, and then accept this one, if you've come to prefer it
$endgroup$
– Learner
Jan 19 at 8:19
add a comment |
$begingroup$
Notice $ sqrt a = r + ssqrt b $ follows from $ c + sqrt{a} = dfrac{e}{d+sqrt b},$ by rationalizing the denominator.
Squaring $, a = r^2!+s^2b + 2rs, sqrt b $ so $ rs = 0,,$ else solving for $sqrt b,$ shows $,sqrt bin Bbb Q$
$sneq 0,$ else $,sqrt a = rin Bbb Q,,$ so $,r=0,$ so $,sqrt a = ssqrt b,$ $overset{largetimessqrt b}Longrightarrowsqrt ab = sbin Bbb Q,,$ contra $a,b$ coprime.
Remark $ $ Above is a special case of the Lemma below, which generalizes to any number of sqrts). See the citations there for generalizations to $n$'th roots.
Lemma $rm [K(sqrt{a},sqrt{b}) : K] = 4 $ if $rm sqrt{a}, sqrt{b}, sqrt{a,b} $ all are not in $rm,K,$ and $rm, 2 ne 0,$ in the field $rm,K.$
Proof $ $ Let $rm L = K(sqrt{b}).,$ Then $rm, [L:K] = 2,$ via $rm,sqrt{b} notin K,,$ so it suffices to prove $rm, [L(sqrt{a}):L] = 2.,$ It fails only if $rm,sqrt{a} in L = K(sqrt{b}), $ and then $rm, sqrt{a} = r + s, sqrt{b} $ for $rm r,sin K.,$ But that's impossible,
since squaring $Rightarrow rm(1)!: a = r^2 + b s^2 + 2,r,s sqrt{b},, $ which contradicts hypotheses as follows:
$rmqquadqquad rs ne 0 Rightarrow sqrt{b} in K $ by solving $(1)$ for $rmsqrt{b},,,$ using $rm,2 ne 0$
$rmqquadqquad s = 0 Rightarrow sqrt{a} in K $ via $rm sqrt{a} = r+s,sqrt b = r in K$
$rmqquadqquad r = 0 Rightarrow sqrt{a,b}in K $ via $rm sqrt{a} = s, sqrt{b}, $times $rm,sqrt{b}quad$
answered Jan 15 at 23:27
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
Notice $ sqrt a = r + ssqrt b $ follows from $ c + sqrt{a} = dfrac{e}{d+sqrt b},$ by rationalizing the denominator.
Squaring $, a = r^2!+s^2b + 2rs, sqrt b $ so $ rs = 0,,$ else solving for $sqrt b,$ shows $,sqrt bin Bbb Q$
$sneq 0,$ else $,sqrt a = rin Bbb Q,,$ so $,r=0,$ so $,sqrt a = ssqrt b,$ $overset{largetimessqrt b}Longrightarrowsqrt ab = sbin Bbb Q,,$ contra $a,b$ coprime.
Remark $ $ Above is a special case of the Lemma below, which generalizes to any number of sqrts). See the citations there for generalizations to $n$'th roots.
Lemma $rm [K(sqrt{a},sqrt{b}) : K] = 4 $ if $rm sqrt{a}, sqrt{b}, sqrt{a,b} $ all are not in $rm,K,$ and $rm, 2 ne 0,$ in the field $rm,K.$
Proof $ $ Let $rm L = K(sqrt{b}).,$ Then $rm, [L:K] = 2,$ via $rm,sqrt{b} notin K,,$ so it suffices to prove $rm, [L(sqrt{a}):L] = 2.,$ It fails only if $rm,sqrt{a} in L = K(sqrt{b}), $ and then $rm, sqrt{a} = r + s, sqrt{b} $ for $rm r,sin K.,$ But that's impossible,
since squaring $Rightarrow rm(1)!: a = r^2 + b s^2 + 2,r,s sqrt{b},, $ which contradicts hypotheses as follows:
$rmqquadqquad rs ne 0 Rightarrow sqrt{b} in K $ by solving $(1)$ for $rmsqrt{b},,,$ using $rm,2 ne 0$
$rmqquadqquad s = 0 Rightarrow sqrt{a} in K $ via $rm sqrt{a} = r+s,sqrt b = r in K$
$rmqquadqquad r = 0 Rightarrow sqrt{a,b}in K $ via $rm sqrt{a} = s, sqrt{b}, $times $rm,sqrt{b}quad$
answered Jan 15 at 23:27
Bill DubuqueBill Dubuque
211k29193646
edited Jan 17 at 3:12
answered Jan 15 at 23:27
Bill DubuqueBill Dubuque
211k29193646
answered Jan 15 at 23:27
Bill DubuqueBill Dubuque
211k29193646
answered Jan 15 at 23:27
Bill DubuqueBill Dubuque
211k29193646
211k29193646
$begingroup$
Now, it's much clearer. This way it looks even simpler, than the one marked as an answer. Keep +1 !
$endgroup$
– cyanide
Jan 17 at 4:00
$begingroup$
@cyanide By clicking on the green tick you may unaccept the currently accepted answer, and then accept this one, if you've come to prefer it
$endgroup$
– Learner
Jan 19 at 8:19
add a comment |
$begingroup$
Now, it's much clearer. This way it looks even simpler, than the one marked as an answer. Keep +1 !
$endgroup$
– cyanide
Jan 17 at 4:00
$begingroup$
@cyanide By clicking on the green tick you may unaccept the currently accepted answer, and then accept this one, if you've come to prefer it
$endgroup$
– Learner
Jan 19 at 8:19
$begingroup$
Now, it's much clearer. This way it looks even simpler, than the one marked as an answer. Keep +1 !
$endgroup$
– cyanide
Jan 17 at 4:00
$begingroup$
Now, it's much clearer. This way it looks even simpler, than the one marked as an answer. Keep +1 !
$endgroup$
– cyanide
Jan 17 at 4:00
$begingroup$
@cyanide By clicking on the green tick you may unaccept the currently accepted answer, and then accept this one, if you've come to prefer it
$endgroup$
– Learner
Jan 19 at 8:19
$begingroup$
@cyanide By clicking on the green tick you may unaccept the currently accepted answer, and then accept this one, if you've come to prefer it
$endgroup$
– Learner
Jan 19 at 8:19
add a comment |
$begingroup$
$(a + sqrt b)(c + sqrt d) = k in mathbb Q$ would mean
$sqrt{b} = frac k{c+sqrt d} - a$
$b = ( frac k{c+sqrt d} - a)^2 in mathbb Q$ which can probably be proven false.
Indeed $( frac k{c+sqrt d} - a)^2 = $
$frac {k(c - sqrt d)}{c^2 - d} -a)^2 =$
$(msqrt d - n)^2$ where $m = frac k{c^2-d}in mathbb Q$ and $n =frac {kc}{c^2 -d} -a in mathbb Q$.
And $(msqrt d -n)^2= m^2d -n^2 - 2nmsqrt d$ which is not rational.
Unless $n$ or $m$ is $0$.
As $kne 0$, $m ne 0$. $n = 0$ if $a = frac {kc}{c^2-d}$.
Hmmm....
answered Jan 15 at 22:51
fleabloodfleablood
71.2k22686
$endgroup$
3
$begingroup$
As for your last comment: e.g. $(-1+sqrt2)(1+sqrt2)=1$.
$endgroup$
– Learner
Jan 15 at 23:10
add a comment |
$begingroup$
$(a + sqrt b)(c + sqrt d) = k in mathbb Q$ would mean
$sqrt{b} = frac k{c+sqrt d} - a$
$b = ( frac k{c+sqrt d} - a)^2 in mathbb Q$ which can probably be proven false.
Indeed $( frac k{c+sqrt d} - a)^2 = $
$frac {k(c - sqrt d)}{c^2 - d} -a)^2 =$
$(msqrt d - n)^2$ where $m = frac k{c^2-d}in mathbb Q$ and $n =frac {kc}{c^2 -d} -a in mathbb Q$.
And $(msqrt d -n)^2= m^2d -n^2 - 2nmsqrt d$ which is not rational.
Unless $n$ or $m$ is $0$.
As $kne 0$, $m ne 0$. $n = 0$ if $a = frac {kc}{c^2-d}$.
Hmmm....
answered Jan 15 at 22:51
fleabloodfleablood
71.2k22686
$endgroup$
3
$begingroup$
As for your last comment: e.g. $(-1+sqrt2)(1+sqrt2)=1$.
$endgroup$
– Learner
Jan 15 at 23:10
add a comment |
$begingroup$
$(a + sqrt b)(c + sqrt d) = k in mathbb Q$ would mean
$sqrt{b} = frac k{c+sqrt d} - a$
$b = ( frac k{c+sqrt d} - a)^2 in mathbb Q$ which can probably be proven false.
Indeed $( frac k{c+sqrt d} - a)^2 = $
$frac {k(c - sqrt d)}{c^2 - d} -a)^2 =$
$(msqrt d - n)^2$ where $m = frac k{c^2-d}in mathbb Q$ and $n =frac {kc}{c^2 -d} -a in mathbb Q$.
And $(msqrt d -n)^2= m^2d -n^2 - 2nmsqrt d$ which is not rational.
Unless $n$ or $m$ is $0$.
As $kne 0$, $m ne 0$. $n = 0$ if $a = frac {kc}{c^2-d}$.
Hmmm....
answered Jan 15 at 22:51
fleabloodfleablood
71.2k22686
$endgroup$
$(a + sqrt b)(c + sqrt d) = k in mathbb Q$ would mean
$sqrt{b} = frac k{c+sqrt d} - a$
$b = ( frac k{c+sqrt d} - a)^2 in mathbb Q$ which can probably be proven false.
Indeed $( frac k{c+sqrt d} - a)^2 = $
$frac {k(c - sqrt d)}{c^2 - d} -a)^2 =$
$(msqrt d - n)^2$ where $m = frac k{c^2-d}in mathbb Q$ and $n =frac {kc}{c^2 -d} -a in mathbb Q$.
And $(msqrt d -n)^2= m^2d -n^2 - 2nmsqrt d$ which is not rational.
Unless $n$ or $m$ is $0$.
As $kne 0$, $m ne 0$. $n = 0$ if $a = frac {kc}{c^2-d}$.
Hmmm....
answered Jan 15 at 22:51
fleabloodfleablood
71.2k22686
edited Jan 15 at 23:30
answered Jan 15 at 22:51
fleabloodfleablood
71.2k22686
answered Jan 15 at 22:51
fleabloodfleablood
71.2k22686
answered Jan 15 at 22:51
fleabloodfleablood
71.2k22686
71.2k22686
3
$begingroup$
As for your last comment: e.g. $(-1+sqrt2)(1+sqrt2)=1$.
$endgroup$
– Learner
Jan 15 at 23:10
add a comment |
3
$begingroup$
As for your last comment: e.g. $(-1+sqrt2)(1+sqrt2)=1$.
$endgroup$
– Learner
Jan 15 at 23:10
3
3
$begingroup$
As for your last comment: e.g. $(-1+sqrt2)(1+sqrt2)=1$.
$endgroup$
– Learner
Jan 15 at 23:10
$begingroup$
As for your last comment: e.g. $(-1+sqrt2)(1+sqrt2)=1$.
$endgroup$
– Learner
Jan 15 at 23:10
add a comment |
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