$sqrt{98} + 4^{frac{1}{3}}$ is irrational [on hold]
Problem is to prove that $sqrt{98} + 4^{frac{1}{3}}$ is irrational.
I know that $sqrt{98}$ and $4^{frac{1}{3}}$ is irrational, but I don't know how to use it in this problem.
proof-writing radicals irrational-numbers
edited 2 days ago
Michael Rozenberg
97.4k1589188
asked 2 days ago
josfjosf
251216
put on hold as off-topic by Andrés E. Caicedo, amWhy, Jyrki Lahtonen, KReiser, josf yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Jyrki Lahtonen, KReiser, josf
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Problem is to prove that $sqrt{98} + 4^{frac{1}{3}}$ is irrational.
I know that $sqrt{98}$ and $4^{frac{1}{3}}$ is irrational, but I don't know how to use it in this problem.
proof-writing radicals irrational-numbers
edited 2 days ago
Michael Rozenberg
97.4k1589188
asked 2 days ago
josfjosf
251216
put on hold as off-topic by Andrés E. Caicedo, amWhy, Jyrki Lahtonen, KReiser, josf yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Jyrki Lahtonen, KReiser, josf
If this question can be reworded to fit the rules in the help center, please edit the question.
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
– SmileyCraft
2 days ago
2
math.stackexchange.com/q/1595280/73324
– vadim123
2 days ago
That's an algebraic integer. An algebraic integer which is in $Bbb Q$ must also be in $Bbb Z$. Your calculator will tell you that's not in $Bbb Z$.
– Lord Shark the Unknown
2 days ago
add a comment |
Problem is to prove that $sqrt{98} + 4^{frac{1}{3}}$ is irrational.
I know that $sqrt{98}$ and $4^{frac{1}{3}}$ is irrational, but I don't know how to use it in this problem.
proof-writing radicals irrational-numbers
edited 2 days ago
Michael Rozenberg
97.4k1589188
asked 2 days ago
josfjosf
251216
Problem is to prove that $sqrt{98} + 4^{frac{1}{3}}$ is irrational.
I know that $sqrt{98}$ and $4^{frac{1}{3}}$ is irrational, but I don't know how to use it in this problem.
proof-writing radicals irrational-numbers
proof-writing radicals irrational-numbers
edited 2 days ago
Michael Rozenberg
97.4k1589188
asked 2 days ago
josfjosf
251216
edited 2 days ago
Michael Rozenberg
97.4k1589188
asked 2 days ago
josfjosf
251216
edited 2 days ago
Michael Rozenberg
97.4k1589188
edited 2 days ago
Michael Rozenberg
97.4k1589188
edited 2 days ago
Michael Rozenberg
97.4k1589188
97.4k1589188
asked 2 days ago
josfjosf
251216
asked 2 days ago
josfjosf
251216
asked 2 days ago
josfjosf
251216
251216
put on hold as off-topic by Andrés E. Caicedo, amWhy, Jyrki Lahtonen, KReiser, josf yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Jyrki Lahtonen, KReiser, josf
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Andrés E. Caicedo, amWhy, Jyrki Lahtonen, KReiser, josf yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Jyrki Lahtonen, KReiser, josf
If this question can be reworded to fit the rules in the help center, please edit the question.
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
– SmileyCraft
2 days ago
2
math.stackexchange.com/q/1595280/73324
– vadim123
2 days ago
That's an algebraic integer. An algebraic integer which is in $Bbb Q$ must also be in $Bbb Z$. Your calculator will tell you that's not in $Bbb Z$.
– Lord Shark the Unknown
2 days ago
add a comment |
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
– SmileyCraft
2 days ago
2
math.stackexchange.com/q/1595280/73324
– vadim123
2 days ago
That's an algebraic integer. An algebraic integer which is in $Bbb Q$ must also be in $Bbb Z$. Your calculator will tell you that's not in $Bbb Z$.
– Lord Shark the Unknown
2 days ago
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
– SmileyCraft
2 days ago
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
– SmileyCraft
2 days ago
2
2
math.stackexchange.com/q/1595280/73324
– vadim123
2 days ago
math.stackexchange.com/q/1595280/73324
– vadim123
2 days ago
That's an algebraic integer. An algebraic integer which is in $Bbb Q$ must also be in $Bbb Z$. Your calculator will tell you that's not in $Bbb Z$.
– Lord Shark the Unknown
2 days ago
That's an algebraic integer. An algebraic integer which is in $Bbb Q$ must also be in $Bbb Z$. Your calculator will tell you that's not in $Bbb Z$.
– Lord Shark the Unknown
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
Let $$7sqrt2+sqrt[3]4=rinmathbb Q.$$
Thus,
$$4=r^3-21sqrt2r^2+294r-686sqrt2$$ or
$$sqrt2=frac{r^3+294r-4}{7(3r^2+98)},$$ which is a contradiction because $sqrt2notinmathbb Q$..
answered 2 days ago
Michael RozenbergMichael Rozenberg
97.4k1589188
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
– Bill Dubuque
2 days ago
Yes, I saw. Nice!
– Michael Rozenberg
2 days ago
add a comment |
If it's rational $= a$ then cubing $,4^{1/3} = a-7sqrt 2,$ yields the contradiction that $ sqrt2$ or $sqrt{-6}$ is rational, by applying the theorem below with $,b,c = -7,2 $ (where $sqrt din F$ means $ d = f^2$ for some $,fin F)$
Theorem $ $ If $,a,b,cin F$ a field,
$bneq 0,,$ then $,(a+bsqrt{c})^3in F,Rightarrow, sqrt cin F,$ or $,sqrt{-3c}in F$
Proof $ $ Cubing $ (a+bsqrt{c})^3 = b(3a^2+cb^2 ) sqrt c + d = e,$ for $,d,ein F$
When $,3a^2+cb^2 neq 0,$ solving the above for $,sqrt c,$ yields $,sqrt c in F$.
Else $, 9a^2 = -3cb^2,Rightarrow, (3a/b)^2 = -3c,Rightarrow,sqrt{-3c}in F$.
answered 2 days ago
Bill DubuqueBill Dubuque
209k29190630
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $$7sqrt2+sqrt[3]4=rinmathbb Q.$$
Thus,
$$4=r^3-21sqrt2r^2+294r-686sqrt2$$ or
$$sqrt2=frac{r^3+294r-4}{7(3r^2+98)},$$ which is a contradiction because $sqrt2notinmathbb Q$..
answered 2 days ago
Michael RozenbergMichael Rozenberg
97.4k1589188
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
– Bill Dubuque
2 days ago
Yes, I saw. Nice!
– Michael Rozenberg
2 days ago
add a comment |
Let $$7sqrt2+sqrt[3]4=rinmathbb Q.$$
Thus,
$$4=r^3-21sqrt2r^2+294r-686sqrt2$$ or
$$sqrt2=frac{r^3+294r-4}{7(3r^2+98)},$$ which is a contradiction because $sqrt2notinmathbb Q$..
answered 2 days ago
Michael RozenbergMichael Rozenberg
97.4k1589188
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
– Bill Dubuque
2 days ago
Yes, I saw. Nice!
– Michael Rozenberg
2 days ago
add a comment |
Let $$7sqrt2+sqrt[3]4=rinmathbb Q.$$
Thus,
$$4=r^3-21sqrt2r^2+294r-686sqrt2$$ or
$$sqrt2=frac{r^3+294r-4}{7(3r^2+98)},$$ which is a contradiction because $sqrt2notinmathbb Q$..
answered 2 days ago
Michael RozenbergMichael Rozenberg
97.4k1589188
Let $$7sqrt2+sqrt[3]4=rinmathbb Q.$$
Thus,
$$4=r^3-21sqrt2r^2+294r-686sqrt2$$ or
$$sqrt2=frac{r^3+294r-4}{7(3r^2+98)},$$ which is a contradiction because $sqrt2notinmathbb Q$..
answered 2 days ago
Michael RozenbergMichael Rozenberg
97.4k1589188
answered 2 days ago
Michael RozenbergMichael Rozenberg
97.4k1589188
answered 2 days ago
Michael RozenbergMichael Rozenberg
97.4k1589188
answered 2 days ago
Michael RozenbergMichael Rozenberg
97.4k1589188
97.4k1589188
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
– Bill Dubuque
2 days ago
Yes, I saw. Nice!
– Michael Rozenberg
2 days ago
add a comment |
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
– Bill Dubuque
2 days ago
Yes, I saw. Nice!
– Michael Rozenberg
2 days ago
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
– Bill Dubuque
2 days ago
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
– Bill Dubuque
2 days ago
Yes, I saw. Nice!
– Michael Rozenberg
2 days ago
Yes, I saw. Nice!
– Michael Rozenberg
2 days ago
add a comment |
If it's rational $= a$ then cubing $,4^{1/3} = a-7sqrt 2,$ yields the contradiction that $ sqrt2$ or $sqrt{-6}$ is rational, by applying the theorem below with $,b,c = -7,2 $ (where $sqrt din F$ means $ d = f^2$ for some $,fin F)$
Theorem $ $ If $,a,b,cin F$ a field,
$bneq 0,,$ then $,(a+bsqrt{c})^3in F,Rightarrow, sqrt cin F,$ or $,sqrt{-3c}in F$
Proof $ $ Cubing $ (a+bsqrt{c})^3 = b(3a^2+cb^2 ) sqrt c + d = e,$ for $,d,ein F$
When $,3a^2+cb^2 neq 0,$ solving the above for $,sqrt c,$ yields $,sqrt c in F$.
Else $, 9a^2 = -3cb^2,Rightarrow, (3a/b)^2 = -3c,Rightarrow,sqrt{-3c}in F$.
answered 2 days ago
Bill DubuqueBill Dubuque
209k29190630
add a comment |
If it's rational $= a$ then cubing $,4^{1/3} = a-7sqrt 2,$ yields the contradiction that $ sqrt2$ or $sqrt{-6}$ is rational, by applying the theorem below with $,b,c = -7,2 $ (where $sqrt din F$ means $ d = f^2$ for some $,fin F)$
Theorem $ $ If $,a,b,cin F$ a field,
$bneq 0,,$ then $,(a+bsqrt{c})^3in F,Rightarrow, sqrt cin F,$ or $,sqrt{-3c}in F$
Proof $ $ Cubing $ (a+bsqrt{c})^3 = b(3a^2+cb^2 ) sqrt c + d = e,$ for $,d,ein F$
When $,3a^2+cb^2 neq 0,$ solving the above for $,sqrt c,$ yields $,sqrt c in F$.
Else $, 9a^2 = -3cb^2,Rightarrow, (3a/b)^2 = -3c,Rightarrow,sqrt{-3c}in F$.
answered 2 days ago
Bill DubuqueBill Dubuque
209k29190630
add a comment |
If it's rational $= a$ then cubing $,4^{1/3} = a-7sqrt 2,$ yields the contradiction that $ sqrt2$ or $sqrt{-6}$ is rational, by applying the theorem below with $,b,c = -7,2 $ (where $sqrt din F$ means $ d = f^2$ for some $,fin F)$
Theorem $ $ If $,a,b,cin F$ a field,
$bneq 0,,$ then $,(a+bsqrt{c})^3in F,Rightarrow, sqrt cin F,$ or $,sqrt{-3c}in F$
Proof $ $ Cubing $ (a+bsqrt{c})^3 = b(3a^2+cb^2 ) sqrt c + d = e,$ for $,d,ein F$
When $,3a^2+cb^2 neq 0,$ solving the above for $,sqrt c,$ yields $,sqrt c in F$.
Else $, 9a^2 = -3cb^2,Rightarrow, (3a/b)^2 = -3c,Rightarrow,sqrt{-3c}in F$.
answered 2 days ago
Bill DubuqueBill Dubuque
209k29190630
If it's rational $= a$ then cubing $,4^{1/3} = a-7sqrt 2,$ yields the contradiction that $ sqrt2$ or $sqrt{-6}$ is rational, by applying the theorem below with $,b,c = -7,2 $ (where $sqrt din F$ means $ d = f^2$ for some $,fin F)$
Theorem $ $ If $,a,b,cin F$ a field,
$bneq 0,,$ then $,(a+bsqrt{c})^3in F,Rightarrow, sqrt cin F,$ or $,sqrt{-3c}in F$
Proof $ $ Cubing $ (a+bsqrt{c})^3 = b(3a^2+cb^2 ) sqrt c + d = e,$ for $,d,ein F$
When $,3a^2+cb^2 neq 0,$ solving the above for $,sqrt c,$ yields $,sqrt c in F$.
Else $, 9a^2 = -3cb^2,Rightarrow, (3a/b)^2 = -3c,Rightarrow,sqrt{-3c}in F$.
answered 2 days ago
Bill DubuqueBill Dubuque
209k29190630
answered 2 days ago
Bill DubuqueBill Dubuque
209k29190630
answered 2 days ago
Bill DubuqueBill Dubuque
209k29190630
answered 2 days ago
Bill DubuqueBill Dubuque
209k29190630
209k29190630
add a comment |
add a comment |
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
– SmileyCraft
2 days ago
2
math.stackexchange.com/q/1595280/73324
– vadim123
2 days ago
That's an algebraic integer. An algebraic integer which is in $Bbb Q$ must also be in $Bbb Z$. Your calculator will tell you that's not in $Bbb Z$.
– Lord Shark the Unknown
2 days ago