$sqrt{98} + 4^{frac{1}{3}}$ is irrational [closed]
$begingroup$
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 Jan 5 at 18:26
Michael Rozenberg
108k1895200
asked Jan 5 at 18:10
josfjosf
286317
$endgroup$
closed as off-topic by Andrés E. Caicedo, Namaste, Jyrki Lahtonen, KReiser, josf Jan 7 at 7:10
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." – Namaste, 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 |
$begingroup$
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 Jan 5 at 18:26
Michael Rozenberg
108k1895200
asked Jan 5 at 18:10
josfjosf
286317
$endgroup$
closed as off-topic by Andrés E. Caicedo, Namaste, Jyrki Lahtonen, KReiser, josf Jan 7 at 7:10
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." – Namaste, Jyrki Lahtonen, KReiser, josf
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
$endgroup$
– SmileyCraft
Jan 5 at 18:12
2
$begingroup$
math.stackexchange.com/q/1595280/73324
$endgroup$
– vadim123
Jan 5 at 18:21
$begingroup$
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 $Bbb Z$.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 4:36
add a comment |
$begingroup$
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 Jan 5 at 18:26
Michael Rozenberg
108k1895200
asked Jan 5 at 18:10
josfjosf
286317
$endgroup$
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 Jan 5 at 18:26
Michael Rozenberg
108k1895200
asked Jan 5 at 18:10
josfjosf
286317
edited Jan 5 at 18:26
Michael Rozenberg
108k1895200
asked Jan 5 at 18:10
josfjosf
286317
edited Jan 5 at 18:26
Michael Rozenberg
108k1895200
edited Jan 5 at 18:26
Michael Rozenberg
108k1895200
edited Jan 5 at 18:26
Michael Rozenberg
108k1895200
108k1895200
asked Jan 5 at 18:10
josfjosf
286317
asked Jan 5 at 18:10
josfjosf
286317
asked Jan 5 at 18:10
josfjosf
286317
286317
closed as off-topic by Andrés E. Caicedo, Namaste, Jyrki Lahtonen, KReiser, josf Jan 7 at 7:10
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." – Namaste, Jyrki Lahtonen, KReiser, josf
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Andrés E. Caicedo, Namaste, Jyrki Lahtonen, KReiser, josf Jan 7 at 7:10
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." – Namaste, Jyrki Lahtonen, KReiser, josf
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
$endgroup$
– SmileyCraft
Jan 5 at 18:12
2
$begingroup$
math.stackexchange.com/q/1595280/73324
$endgroup$
– vadim123
Jan 5 at 18:21
$begingroup$
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 $Bbb Z$.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 4:36
add a comment |
$begingroup$
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
$endgroup$
– SmileyCraft
Jan 5 at 18:12
2
$begingroup$
math.stackexchange.com/q/1595280/73324
$endgroup$
– vadim123
Jan 5 at 18:21
$begingroup$
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 $Bbb Z$.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 4:36
$begingroup$
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
$endgroup$
– SmileyCraft
Jan 5 at 18:12
$begingroup$
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
$endgroup$
– SmileyCraft
Jan 5 at 18:12
2
2
$begingroup$
math.stackexchange.com/q/1595280/73324
$endgroup$
– vadim123
Jan 5 at 18:21
$begingroup$
math.stackexchange.com/q/1595280/73324
$endgroup$
– vadim123
Jan 5 at 18:21
$begingroup$
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 $Bbb Z$.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 4:36
$begingroup$
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 $Bbb Z$.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 4:36
add a comment |
2 Answers
2
active
oldest
votes
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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 Jan 5 at 18:23
Michael RozenbergMichael Rozenberg
108k1895200
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More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
$endgroup$
– Bill Dubuque
Jan 5 at 20:38
$begingroup$
Yes, I saw. Nice!
$endgroup$
– Michael Rozenberg
Jan 5 at 20:50
add a comment |
$begingroup$
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 Jan 5 at 20:22
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Jan 5 at 18:23
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
$endgroup$
– Bill Dubuque
Jan 5 at 20:38
$begingroup$
Yes, I saw. Nice!
$endgroup$
– Michael Rozenberg
Jan 5 at 20:50
add a comment |
$begingroup$
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 Jan 5 at 18:23
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
$endgroup$
– Bill Dubuque
Jan 5 at 20:38
$begingroup$
Yes, I saw. Nice!
$endgroup$
– Michael Rozenberg
Jan 5 at 20:50
add a comment |
$begingroup$
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 Jan 5 at 18:23
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
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 Jan 5 at 18:23
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 5 at 18:23
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 5 at 18:23
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 5 at 18:23
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
$begingroup$
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
$endgroup$
– Bill Dubuque
Jan 5 at 20:38
$begingroup$
Yes, I saw. Nice!
$endgroup$
– Michael Rozenberg
Jan 5 at 20:50
add a comment |
$begingroup$
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
$endgroup$
– Bill Dubuque
Jan 5 at 20:38
$begingroup$
Yes, I saw. Nice!
$endgroup$
– Michael Rozenberg
Jan 5 at 20:50
$begingroup$
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
$endgroup$
– Bill Dubuque
Jan 5 at 20:38
$begingroup$
More generally it's true in any field where $2$ and $-6$ are not squares - see my answer.
$endgroup$
– Bill Dubuque
Jan 5 at 20:38
$begingroup$
Yes, I saw. Nice!
$endgroup$
– Michael Rozenberg
Jan 5 at 20:50
$begingroup$
Yes, I saw. Nice!
$endgroup$
– Michael Rozenberg
Jan 5 at 20:50
add a comment |
$begingroup$
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 Jan 5 at 20:22
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 20:22
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 20:22
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
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 Jan 5 at 20:22
Bill DubuqueBill Dubuque
212k29195654
answered Jan 5 at 20:22
Bill DubuqueBill Dubuque
212k29195654
answered Jan 5 at 20:22
Bill DubuqueBill Dubuque
212k29195654
answered Jan 5 at 20:22
Bill DubuqueBill Dubuque
212k29195654
212k29195654
add a comment |
add a comment |
$begingroup$
Do you know the rational root theorem? You should check out Mathologer's latest video. youtube.com/watch?v=D6AFxJdJYW4
$endgroup$
– SmileyCraft
Jan 5 at 18:12
2
$begingroup$
math.stackexchange.com/q/1595280/73324
$endgroup$
– vadim123
Jan 5 at 18:21
$begingroup$
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 $Bbb Z$.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 4:36