Proof t is irrational t = a-bs ,Given a and b are rational numbers, b $neq 0$ and s is irrational.
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Proof $t$ is irrational $ t = a-bs $ , Given $a$ and $b$ are rational numbers, $b neq 0$ and $s$ is irrational.
Hence show that $(sqrt3-1)/(sqrt3+1)$ is irrational
real-analysis abstract-algebra
asked Nov 17 at 12:42
Eric kioko
265
add a comment |
up vote
2
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Proof $t$ is irrational $ t = a-bs $ , Given $a$ and $b$ are rational numbers, $b neq 0$ and $s$ is irrational.
Hence show that $(sqrt3-1)/(sqrt3+1)$ is irrational
real-analysis abstract-algebra
asked Nov 17 at 12:42
Eric kioko
265
Can you simplify $(sqrt3-1)/(sqrt3+1)$?
– Lord Shark the Unknown
Nov 17 at 12:52
The problem is in two parts, and it's not clear to me whether you are saying that you have already solved the first part or not.
– TonyK
Nov 17 at 12:55
Try to prove that $t$ is irrational by contradiction i.e. assume that $t$ was a rational number.
– clark
Nov 17 at 13:00
@TonyK I got stuck not sure. $t=a-bs$ I assumed t to be rational i.e $p/q$ where $p$ and $q$ member of integers and $pneq 0$ thus $p/(a-bs)=q$ So p is divisible by $a-bs $ but don't know how to find a divisor for $q$ to disqualify $t$ as rational
– Eric kioko
Nov 17 at 14:15
You have not used the first part to prove the second part. Rather, you have used the same method of proof. But you don't need to go through that again! You can just put $a=2$, $b=1$, and $s=sqrt 3$, and apply what you proved in the first part.
– TonyK
Nov 17 at 15:23
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Proof $t$ is irrational $ t = a-bs $ , Given $a$ and $b$ are rational numbers, $b neq 0$ and $s$ is irrational.
Hence show that $(sqrt3-1)/(sqrt3+1)$ is irrational
real-analysis abstract-algebra
asked Nov 17 at 12:42
Eric kioko
265
Proof $t$ is irrational $ t = a-bs $ , Given $a$ and $b$ are rational numbers, $b neq 0$ and $s$ is irrational.
Hence show that $(sqrt3-1)/(sqrt3+1)$ is irrational
real-analysis abstract-algebra
real-analysis abstract-algebra
asked Nov 17 at 12:42
Eric kioko
265
asked Nov 17 at 12:42
Eric kioko
265
edited Nov 17 at 15:44
asked Nov 17 at 12:42
Eric kioko
265
asked Nov 17 at 12:42
Eric kioko
265
asked Nov 17 at 12:42
Eric kioko
265
265
Can you simplify $(sqrt3-1)/(sqrt3+1)$?
– Lord Shark the Unknown
Nov 17 at 12:52
The problem is in two parts, and it's not clear to me whether you are saying that you have already solved the first part or not.
– TonyK
Nov 17 at 12:55
Try to prove that $t$ is irrational by contradiction i.e. assume that $t$ was a rational number.
– clark
Nov 17 at 13:00
@TonyK I got stuck not sure. $t=a-bs$ I assumed t to be rational i.e $p/q$ where $p$ and $q$ member of integers and $pneq 0$ thus $p/(a-bs)=q$ So p is divisible by $a-bs $ but don't know how to find a divisor for $q$ to disqualify $t$ as rational
– Eric kioko
Nov 17 at 14:15
You have not used the first part to prove the second part. Rather, you have used the same method of proof. But you don't need to go through that again! You can just put $a=2$, $b=1$, and $s=sqrt 3$, and apply what you proved in the first part.
– TonyK
Nov 17 at 15:23
add a comment |
Can you simplify $(sqrt3-1)/(sqrt3+1)$?
– Lord Shark the Unknown
Nov 17 at 12:52
The problem is in two parts, and it's not clear to me whether you are saying that you have already solved the first part or not.
– TonyK
Nov 17 at 12:55
Try to prove that $t$ is irrational by contradiction i.e. assume that $t$ was a rational number.
– clark
Nov 17 at 13:00
@TonyK I got stuck not sure. $t=a-bs$ I assumed t to be rational i.e $p/q$ where $p$ and $q$ member of integers and $pneq 0$ thus $p/(a-bs)=q$ So p is divisible by $a-bs $ but don't know how to find a divisor for $q$ to disqualify $t$ as rational
– Eric kioko
Nov 17 at 14:15
You have not used the first part to prove the second part. Rather, you have used the same method of proof. But you don't need to go through that again! You can just put $a=2$, $b=1$, and $s=sqrt 3$, and apply what you proved in the first part.
– TonyK
Nov 17 at 15:23
Can you simplify $(sqrt3-1)/(sqrt3+1)$?
– Lord Shark the Unknown
Nov 17 at 12:52
Can you simplify $(sqrt3-1)/(sqrt3+1)$?
– Lord Shark the Unknown
Nov 17 at 12:52
The problem is in two parts, and it's not clear to me whether you are saying that you have already solved the first part or not.
– TonyK
Nov 17 at 12:55
The problem is in two parts, and it's not clear to me whether you are saying that you have already solved the first part or not.
– TonyK
Nov 17 at 12:55
Try to prove that $t$ is irrational by contradiction i.e. assume that $t$ was a rational number.
– clark
Nov 17 at 13:00
Try to prove that $t$ is irrational by contradiction i.e. assume that $t$ was a rational number.
– clark
Nov 17 at 13:00
@TonyK I got stuck not sure. $t=a-bs$ I assumed t to be rational i.e $p/q$ where $p$ and $q$ member of integers and $pneq 0$ thus $p/(a-bs)=q$ So p is divisible by $a-bs $ but don't know how to find a divisor for $q$ to disqualify $t$ as rational
– Eric kioko
Nov 17 at 14:15
@TonyK I got stuck not sure. $t=a-bs$ I assumed t to be rational i.e $p/q$ where $p$ and $q$ member of integers and $pneq 0$ thus $p/(a-bs)=q$ So p is divisible by $a-bs $ but don't know how to find a divisor for $q$ to disqualify $t$ as rational
– Eric kioko
Nov 17 at 14:15
You have not used the first part to prove the second part. Rather, you have used the same method of proof. But you don't need to go through that again! You can just put $a=2$, $b=1$, and $s=sqrt 3$, and apply what you proved in the first part.
– TonyK
Nov 17 at 15:23
You have not used the first part to prove the second part. Rather, you have used the same method of proof. But you don't need to go through that again! You can just put $a=2$, $b=1$, and $s=sqrt 3$, and apply what you proved in the first part.
– TonyK
Nov 17 at 15:23
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
For the first part: in a comment you have solved for $s$ in terms of $a$, $b$ and $t$. You know the first two are rational. What could you say about $s$ if $t$ were rational too?
You have started the second part correctly. Now it's in a form where you can apply the first part.
answered Nov 17 at 14:35
Ethan Bolker
39.7k543103
Thanks for the eye opener, I have attempted check to see if I'm right.
– Eric kioko
Nov 17 at 15:14
WIth your eyes open you have a pretty much correct solution. You need the fact that the rationals are closed under division, too, which you use but don't state. For part 2 you have given a direct proof based on the idea from part 1, but didn't explicitly use that result, so the "hence" in the question hasn't been addressed. You could edit the question to remove the solution and then post the solution as an answer to your own question. That's allowed here, and it will take the question off the unanswered queue.
– Ethan Bolker
Nov 17 at 15:18
add a comment |
up vote
1
down vote
accepted
For the first part
$a/b -s = t/b$
$(a-t)/b=s$
Assuming t was rational would mean $ s $ is rational since rational numbers are closed under addition and subtraction.
This contradicts s being irrational.
Therefore t is irrational.
The second part
$(surd3-1)/(surd3+1) * (surd3-1)/(surd3-1) $
$=2-surd3$
substituting $2-surd3$
in $t=a-bs$
$a=2,b=1,s=surd3$
$t=2-surd3$
Therefore $t=2-surd3$ is irrational too.
answered Nov 17 at 15:49
Eric kioko
265
You can avoid the checks: $2-sqrt{3}$ is irrational because $s=sqrt{3}$ is irrational and you can take $a=2$, $b=1$.
– egreg
Nov 17 at 16:56
@egreg Thanks, that's how its supposed to be done. I'm gonna edit my answer to reflect that.
– Eric kioko
Nov 21 at 18:53
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
For the first part: in a comment you have solved for $s$ in terms of $a$, $b$ and $t$. You know the first two are rational. What could you say about $s$ if $t$ were rational too?
You have started the second part correctly. Now it's in a form where you can apply the first part.
answered Nov 17 at 14:35
Ethan Bolker
39.7k543103
Thanks for the eye opener, I have attempted check to see if I'm right.
– Eric kioko
Nov 17 at 15:14
WIth your eyes open you have a pretty much correct solution. You need the fact that the rationals are closed under division, too, which you use but don't state. For part 2 you have given a direct proof based on the idea from part 1, but didn't explicitly use that result, so the "hence" in the question hasn't been addressed. You could edit the question to remove the solution and then post the solution as an answer to your own question. That's allowed here, and it will take the question off the unanswered queue.
– Ethan Bolker
Nov 17 at 15:18
add a comment |
up vote
2
down vote
For the first part: in a comment you have solved for $s$ in terms of $a$, $b$ and $t$. You know the first two are rational. What could you say about $s$ if $t$ were rational too?
You have started the second part correctly. Now it's in a form where you can apply the first part.
answered Nov 17 at 14:35
Ethan Bolker
39.7k543103
Thanks for the eye opener, I have attempted check to see if I'm right.
– Eric kioko
Nov 17 at 15:14
WIth your eyes open you have a pretty much correct solution. You need the fact that the rationals are closed under division, too, which you use but don't state. For part 2 you have given a direct proof based on the idea from part 1, but didn't explicitly use that result, so the "hence" in the question hasn't been addressed. You could edit the question to remove the solution and then post the solution as an answer to your own question. That's allowed here, and it will take the question off the unanswered queue.
– Ethan Bolker
Nov 17 at 15:18
add a comment |
up vote
2
down vote
up vote
2
down vote
For the first part: in a comment you have solved for $s$ in terms of $a$, $b$ and $t$. You know the first two are rational. What could you say about $s$ if $t$ were rational too?
You have started the second part correctly. Now it's in a form where you can apply the first part.
answered Nov 17 at 14:35
Ethan Bolker
39.7k543103
For the first part: in a comment you have solved for $s$ in terms of $a$, $b$ and $t$. You know the first two are rational. What could you say about $s$ if $t$ were rational too?
You have started the second part correctly. Now it's in a form where you can apply the first part.
answered Nov 17 at 14:35
Ethan Bolker
39.7k543103
answered Nov 17 at 14:35
Ethan Bolker
39.7k543103
answered Nov 17 at 14:35
Ethan Bolker
39.7k543103
answered Nov 17 at 14:35
Ethan Bolker
39.7k543103
39.7k543103
Thanks for the eye opener, I have attempted check to see if I'm right.
– Eric kioko
Nov 17 at 15:14
WIth your eyes open you have a pretty much correct solution. You need the fact that the rationals are closed under division, too, which you use but don't state. For part 2 you have given a direct proof based on the idea from part 1, but didn't explicitly use that result, so the "hence" in the question hasn't been addressed. You could edit the question to remove the solution and then post the solution as an answer to your own question. That's allowed here, and it will take the question off the unanswered queue.
– Ethan Bolker
Nov 17 at 15:18
add a comment |
Thanks for the eye opener, I have attempted check to see if I'm right.
– Eric kioko
Nov 17 at 15:14
WIth your eyes open you have a pretty much correct solution. You need the fact that the rationals are closed under division, too, which you use but don't state. For part 2 you have given a direct proof based on the idea from part 1, but didn't explicitly use that result, so the "hence" in the question hasn't been addressed. You could edit the question to remove the solution and then post the solution as an answer to your own question. That's allowed here, and it will take the question off the unanswered queue.
– Ethan Bolker
Nov 17 at 15:18
Thanks for the eye opener, I have attempted check to see if I'm right.
– Eric kioko
Nov 17 at 15:14
Thanks for the eye opener, I have attempted check to see if I'm right.
– Eric kioko
Nov 17 at 15:14
WIth your eyes open you have a pretty much correct solution. You need the fact that the rationals are closed under division, too, which you use but don't state. For part 2 you have given a direct proof based on the idea from part 1, but didn't explicitly use that result, so the "hence" in the question hasn't been addressed. You could edit the question to remove the solution and then post the solution as an answer to your own question. That's allowed here, and it will take the question off the unanswered queue.
– Ethan Bolker
Nov 17 at 15:18
WIth your eyes open you have a pretty much correct solution. You need the fact that the rationals are closed under division, too, which you use but don't state. For part 2 you have given a direct proof based on the idea from part 1, but didn't explicitly use that result, so the "hence" in the question hasn't been addressed. You could edit the question to remove the solution and then post the solution as an answer to your own question. That's allowed here, and it will take the question off the unanswered queue.
– Ethan Bolker
Nov 17 at 15:18
add a comment |
up vote
1
down vote
accepted
For the first part
$a/b -s = t/b$
$(a-t)/b=s$
Assuming t was rational would mean $ s $ is rational since rational numbers are closed under addition and subtraction.
This contradicts s being irrational.
Therefore t is irrational.
The second part
$(surd3-1)/(surd3+1) * (surd3-1)/(surd3-1) $
$=2-surd3$
substituting $2-surd3$
in $t=a-bs$
$a=2,b=1,s=surd3$
$t=2-surd3$
Therefore $t=2-surd3$ is irrational too.
answered Nov 17 at 15:49
Eric kioko
265
You can avoid the checks: $2-sqrt{3}$ is irrational because $s=sqrt{3}$ is irrational and you can take $a=2$, $b=1$.
– egreg
Nov 17 at 16:56
@egreg Thanks, that's how its supposed to be done. I'm gonna edit my answer to reflect that.
– Eric kioko
Nov 21 at 18:53
add a comment |
up vote
1
down vote
accepted
For the first part
$a/b -s = t/b$
$(a-t)/b=s$
Assuming t was rational would mean $ s $ is rational since rational numbers are closed under addition and subtraction.
This contradicts s being irrational.
Therefore t is irrational.
The second part
$(surd3-1)/(surd3+1) * (surd3-1)/(surd3-1) $
$=2-surd3$
substituting $2-surd3$
in $t=a-bs$
$a=2,b=1,s=surd3$
$t=2-surd3$
Therefore $t=2-surd3$ is irrational too.
answered Nov 17 at 15:49
Eric kioko
265
You can avoid the checks: $2-sqrt{3}$ is irrational because $s=sqrt{3}$ is irrational and you can take $a=2$, $b=1$.
– egreg
Nov 17 at 16:56
@egreg Thanks, that's how its supposed to be done. I'm gonna edit my answer to reflect that.
– Eric kioko
Nov 21 at 18:53
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For the first part
$a/b -s = t/b$
$(a-t)/b=s$
Assuming t was rational would mean $ s $ is rational since rational numbers are closed under addition and subtraction.
This contradicts s being irrational.
Therefore t is irrational.
The second part
$(surd3-1)/(surd3+1) * (surd3-1)/(surd3-1) $
$=2-surd3$
substituting $2-surd3$
in $t=a-bs$
$a=2,b=1,s=surd3$
$t=2-surd3$
Therefore $t=2-surd3$ is irrational too.
answered Nov 17 at 15:49
Eric kioko
265
For the first part
$a/b -s = t/b$
$(a-t)/b=s$
Assuming t was rational would mean $ s $ is rational since rational numbers are closed under addition and subtraction.
This contradicts s being irrational.
Therefore t is irrational.
The second part
$(surd3-1)/(surd3+1) * (surd3-1)/(surd3-1) $
$=2-surd3$
substituting $2-surd3$
in $t=a-bs$
$a=2,b=1,s=surd3$
$t=2-surd3$
Therefore $t=2-surd3$ is irrational too.
answered Nov 17 at 15:49
Eric kioko
265
edited Nov 25 at 8:09
answered Nov 17 at 15:49
Eric kioko
265
answered Nov 17 at 15:49
Eric kioko
265
answered Nov 17 at 15:49
Eric kioko
265
265
You can avoid the checks: $2-sqrt{3}$ is irrational because $s=sqrt{3}$ is irrational and you can take $a=2$, $b=1$.
– egreg
Nov 17 at 16:56
@egreg Thanks, that's how its supposed to be done. I'm gonna edit my answer to reflect that.
– Eric kioko
Nov 21 at 18:53
add a comment |
You can avoid the checks: $2-sqrt{3}$ is irrational because $s=sqrt{3}$ is irrational and you can take $a=2$, $b=1$.
– egreg
Nov 17 at 16:56
@egreg Thanks, that's how its supposed to be done. I'm gonna edit my answer to reflect that.
– Eric kioko
Nov 21 at 18:53
You can avoid the checks: $2-sqrt{3}$ is irrational because $s=sqrt{3}$ is irrational and you can take $a=2$, $b=1$.
– egreg
Nov 17 at 16:56
You can avoid the checks: $2-sqrt{3}$ is irrational because $s=sqrt{3}$ is irrational and you can take $a=2$, $b=1$.
– egreg
Nov 17 at 16:56
@egreg Thanks, that's how its supposed to be done. I'm gonna edit my answer to reflect that.
– Eric kioko
Nov 21 at 18:53
@egreg Thanks, that's how its supposed to be done. I'm gonna edit my answer to reflect that.
– Eric kioko
Nov 21 at 18:53
add a comment |
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Can you simplify $(sqrt3-1)/(sqrt3+1)$?
– Lord Shark the Unknown
Nov 17 at 12:52
The problem is in two parts, and it's not clear to me whether you are saying that you have already solved the first part or not.
– TonyK
Nov 17 at 12:55
Try to prove that $t$ is irrational by contradiction i.e. assume that $t$ was a rational number.
– clark
Nov 17 at 13:00
@TonyK I got stuck not sure. $t=a-bs$ I assumed t to be rational i.e $p/q$ where $p$ and $q$ member of integers and $pneq 0$ thus $p/(a-bs)=q$ So p is divisible by $a-bs $ but don't know how to find a divisor for $q$ to disqualify $t$ as rational
– Eric kioko
Nov 17 at 14:15
You have not used the first part to prove the second part. Rather, you have used the same method of proof. But you don't need to go through that again! You can just put $a=2$, $b=1$, and $s=sqrt 3$, and apply what you proved in the first part.
– TonyK
Nov 17 at 15:23