Prove of irrationality [closed]
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how can I prove that $(1-sqrt{2})^z $ is never rational for any integer $z$ different from $0$ ?
algebra-precalculus irrational-numbers
edited Nov 16 at 23:38
Servaes
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asked Nov 16 at 18:45
Lance
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closed as off-topic by T. Bongers, Servaes, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Nosrati Nov 16 at 19:03
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Servaes, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Nosrati
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how can I prove that $(1-sqrt{2})^z $ is never rational for any integer $z$ different from $0$ ?
algebra-precalculus irrational-numbers
edited Nov 16 at 23:38
Servaes
20.9k33792
asked Nov 16 at 18:45
Lance
5910
closed as off-topic by T. Bongers, Servaes, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Nosrati Nov 16 at 19:03
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Servaes, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Nosrati
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
how can I prove that $(1-sqrt{2})^z $ is never rational for any integer $z$ different from $0$ ?
algebra-precalculus irrational-numbers
edited Nov 16 at 23:38
Servaes
20.9k33792
asked Nov 16 at 18:45
Lance
5910
how can I prove that $(1-sqrt{2})^z $ is never rational for any integer $z$ different from $0$ ?
algebra-precalculus irrational-numbers
algebra-precalculus irrational-numbers
edited Nov 16 at 23:38
Servaes
20.9k33792
asked Nov 16 at 18:45
Lance
5910
edited Nov 16 at 23:38
Servaes
20.9k33792
asked Nov 16 at 18:45
Lance
5910
edited Nov 16 at 23:38
Servaes
20.9k33792
edited Nov 16 at 23:38
Servaes
20.9k33792
edited Nov 16 at 23:38
Servaes
20.9k33792
20.9k33792
asked Nov 16 at 18:45
Lance
5910
asked Nov 16 at 18:45
Lance
5910
asked Nov 16 at 18:45
Lance
5910
5910
closed as off-topic by T. Bongers, Servaes, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Nosrati Nov 16 at 19:03
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Servaes, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Nosrati
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by T. Bongers, Servaes, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Nosrati Nov 16 at 19:03
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Servaes, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Nosrati
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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Note that $(1-sqrt{2})(1+sqrt{2})=-1$ so $(1-sqrt{2})^z$ is rational if and only if
$$(1+sqrt{2})^z=sum_{k=0}^zbinom{z}{k}sqrt{2}^k,$$
is rational. But this is clearly of the form $a+bsqrt{2}$ for some positive integers $a$ and $b$.
answered Nov 16 at 18:55
Servaes
20.9k33792
2
Why go through $(1+sqrt 2)^z$? The binomial theorem works just as well on $(1-sqrt2)^z$ directly (yes, $b$ becomes negative, but that doesn't change the result in the slightest).
– Arthur
Nov 16 at 18:58
@Arthur I think in the $(1-sqrt{2})^z$ case we will need to prove that $b ne 0$
– user1952500
Nov 16 at 18:59
All terms with $sqrt2$ raised to some odd power will have negative sign. So you get the exact same $b$ as you do for $(1+sqrt2)^z$, only with opposite sign.
– Arthur
Nov 16 at 19:03
@Arthur You are absolutely right, I just found that this representation illustrates the point more clearly.
– Servaes
Nov 16 at 19:08
add a comment |
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Note that $(1-sqrt{2})(1+sqrt{2})=-1$ so $(1-sqrt{2})^z$ is rational if and only if
$$(1+sqrt{2})^z=sum_{k=0}^zbinom{z}{k}sqrt{2}^k,$$
is rational. But this is clearly of the form $a+bsqrt{2}$ for some positive integers $a$ and $b$.
answered Nov 16 at 18:55
Servaes
20.9k33792
2
Why go through $(1+sqrt 2)^z$? The binomial theorem works just as well on $(1-sqrt2)^z$ directly (yes, $b$ becomes negative, but that doesn't change the result in the slightest).
– Arthur
Nov 16 at 18:58
@Arthur I think in the $(1-sqrt{2})^z$ case we will need to prove that $b ne 0$
– user1952500
Nov 16 at 18:59
All terms with $sqrt2$ raised to some odd power will have negative sign. So you get the exact same $b$ as you do for $(1+sqrt2)^z$, only with opposite sign.
– Arthur
Nov 16 at 19:03
@Arthur You are absolutely right, I just found that this representation illustrates the point more clearly.
– Servaes
Nov 16 at 19:08
add a comment |
up vote
3
down vote
Note that $(1-sqrt{2})(1+sqrt{2})=-1$ so $(1-sqrt{2})^z$ is rational if and only if
$$(1+sqrt{2})^z=sum_{k=0}^zbinom{z}{k}sqrt{2}^k,$$
is rational. But this is clearly of the form $a+bsqrt{2}$ for some positive integers $a$ and $b$.
answered Nov 16 at 18:55
Servaes
20.9k33792
2
Why go through $(1+sqrt 2)^z$? The binomial theorem works just as well on $(1-sqrt2)^z$ directly (yes, $b$ becomes negative, but that doesn't change the result in the slightest).
– Arthur
Nov 16 at 18:58
@Arthur I think in the $(1-sqrt{2})^z$ case we will need to prove that $b ne 0$
– user1952500
Nov 16 at 18:59
All terms with $sqrt2$ raised to some odd power will have negative sign. So you get the exact same $b$ as you do for $(1+sqrt2)^z$, only with opposite sign.
– Arthur
Nov 16 at 19:03
@Arthur You are absolutely right, I just found that this representation illustrates the point more clearly.
– Servaes
Nov 16 at 19:08
add a comment |
up vote
3
down vote
up vote
3
down vote
Note that $(1-sqrt{2})(1+sqrt{2})=-1$ so $(1-sqrt{2})^z$ is rational if and only if
$$(1+sqrt{2})^z=sum_{k=0}^zbinom{z}{k}sqrt{2}^k,$$
is rational. But this is clearly of the form $a+bsqrt{2}$ for some positive integers $a$ and $b$.
answered Nov 16 at 18:55
Servaes
20.9k33792
Note that $(1-sqrt{2})(1+sqrt{2})=-1$ so $(1-sqrt{2})^z$ is rational if and only if
$$(1+sqrt{2})^z=sum_{k=0}^zbinom{z}{k}sqrt{2}^k,$$
is rational. But this is clearly of the form $a+bsqrt{2}$ for some positive integers $a$ and $b$.
answered Nov 16 at 18:55
Servaes
20.9k33792
answered Nov 16 at 18:55
Servaes
20.9k33792
answered Nov 16 at 18:55
Servaes
20.9k33792
answered Nov 16 at 18:55
Servaes
20.9k33792
20.9k33792
2
Why go through $(1+sqrt 2)^z$? The binomial theorem works just as well on $(1-sqrt2)^z$ directly (yes, $b$ becomes negative, but that doesn't change the result in the slightest).
– Arthur
Nov 16 at 18:58
@Arthur I think in the $(1-sqrt{2})^z$ case we will need to prove that $b ne 0$
– user1952500
Nov 16 at 18:59
All terms with $sqrt2$ raised to some odd power will have negative sign. So you get the exact same $b$ as you do for $(1+sqrt2)^z$, only with opposite sign.
– Arthur
Nov 16 at 19:03
@Arthur You are absolutely right, I just found that this representation illustrates the point more clearly.
– Servaes
Nov 16 at 19:08
add a comment |
2
Why go through $(1+sqrt 2)^z$? The binomial theorem works just as well on $(1-sqrt2)^z$ directly (yes, $b$ becomes negative, but that doesn't change the result in the slightest).
– Arthur
Nov 16 at 18:58
@Arthur I think in the $(1-sqrt{2})^z$ case we will need to prove that $b ne 0$
– user1952500
Nov 16 at 18:59
All terms with $sqrt2$ raised to some odd power will have negative sign. So you get the exact same $b$ as you do for $(1+sqrt2)^z$, only with opposite sign.
– Arthur
Nov 16 at 19:03
@Arthur You are absolutely right, I just found that this representation illustrates the point more clearly.
– Servaes
Nov 16 at 19:08
2
2
Why go through $(1+sqrt 2)^z$? The binomial theorem works just as well on $(1-sqrt2)^z$ directly (yes, $b$ becomes negative, but that doesn't change the result in the slightest).
– Arthur
Nov 16 at 18:58
Why go through $(1+sqrt 2)^z$? The binomial theorem works just as well on $(1-sqrt2)^z$ directly (yes, $b$ becomes negative, but that doesn't change the result in the slightest).
– Arthur
Nov 16 at 18:58
@Arthur I think in the $(1-sqrt{2})^z$ case we will need to prove that $b ne 0$
– user1952500
Nov 16 at 18:59
@Arthur I think in the $(1-sqrt{2})^z$ case we will need to prove that $b ne 0$
– user1952500
Nov 16 at 18:59
All terms with $sqrt2$ raised to some odd power will have negative sign. So you get the exact same $b$ as you do for $(1+sqrt2)^z$, only with opposite sign.
– Arthur
Nov 16 at 19:03
All terms with $sqrt2$ raised to some odd power will have negative sign. So you get the exact same $b$ as you do for $(1+sqrt2)^z$, only with opposite sign.
– Arthur
Nov 16 at 19:03
@Arthur You are absolutely right, I just found that this representation illustrates the point more clearly.
– Servaes
Nov 16 at 19:08
@Arthur You are absolutely right, I just found that this representation illustrates the point more clearly.
– Servaes
Nov 16 at 19:08
add a comment |
