Proof By Contradiction With Rational and Irrational Numbers
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
The question I am working on is:
Use a proof by contradiction to prove that the sum of an irrational number and a rational number is irrational.
After searching through Google, to see if this particular question had been asked before, I found this: http://answers.yahoo.com/question/index?qid=20081012182747AA3AaHz
I am looking exclusively at Charles' answer. I understand the proof, and its steps, I just don't understand how it proves the original statement is true; I only see it as proving that, if you add two rational numbers, you get another rational number. Could someone help me?
discrete-mathematics proof-writing
asked Feb 14 '13 at 13:24
Mack
1,9611462114
add a comment |Â
up vote
3
down vote
favorite
The question I am working on is:
Use a proof by contradiction to prove that the sum of an irrational number and a rational number is irrational.
After searching through Google, to see if this particular question had been asked before, I found this: http://answers.yahoo.com/question/index?qid=20081012182747AA3AaHz
I am looking exclusively at Charles' answer. I understand the proof, and its steps, I just don't understand how it proves the original statement is true; I only see it as proving that, if you add two rational numbers, you get another rational number. Could someone help me?
discrete-mathematics proof-writing
asked Feb 14 '13 at 13:24
Mack
1,9611462114
1
Do you understand proof by contradiction?
– Thomas Andrews
Feb 14 '13 at 13:31
@ThomasAndrews Yes, I do. Essentially, the idea is that you proof that something, say $q$, implies something that is, say $p$, and something that isn't, say $neg p$. Hence, $q implies (p wedge neg p)$ Which is absurd: something can't both be and not be at the same time--unless, of course, we are discussing quantum mechanics.
– Mack
Feb 14 '13 at 13:37
Okay, now define your $p$ and $q$.
– Joe Z.
Feb 14 '13 at 13:39
Well, I know that $p$ would definitely be that "$i$ is irrational"; however, I am not sure exactly what $q$ would be. Would $q$ be "the sum of a rational and irrational number?"
– Mack
Feb 14 '13 at 13:42
@Mack q⟹(p∧¬p) isn't saying that something is:both be and not be at the same timeFor instance if p is true then ¬p is false, it's not saying that p=¬p, such would be false which is the same logic here: p∧¬p. Ultimately they're not the same, and they are logical opposites.
– CTS_AE
May 17 '17 at 4:38
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The question I am working on is:
Use a proof by contradiction to prove that the sum of an irrational number and a rational number is irrational.
After searching through Google, to see if this particular question had been asked before, I found this: http://answers.yahoo.com/question/index?qid=20081012182747AA3AaHz
I am looking exclusively at Charles' answer. I understand the proof, and its steps, I just don't understand how it proves the original statement is true; I only see it as proving that, if you add two rational numbers, you get another rational number. Could someone help me?
discrete-mathematics proof-writing
asked Feb 14 '13 at 13:24
Mack
1,9611462114
The question I am working on is:
Use a proof by contradiction to prove that the sum of an irrational number and a rational number is irrational.
After searching through Google, to see if this particular question had been asked before, I found this: http://answers.yahoo.com/question/index?qid=20081012182747AA3AaHz
I am looking exclusively at Charles' answer. I understand the proof, and its steps, I just don't understand how it proves the original statement is true; I only see it as proving that, if you add two rational numbers, you get another rational number. Could someone help me?
discrete-mathematics proof-writing
asked Feb 14 '13 at 13:24
Mack
1,9611462114
asked Feb 14 '13 at 13:24
Mack
1,9611462114
asked Feb 14 '13 at 13:24
Mack
1,9611462114
asked Feb 14 '13 at 13:24
Mack
1,9611462114
1,9611462114
1
Do you understand proof by contradiction?
– Thomas Andrews
Feb 14 '13 at 13:31
@ThomasAndrews Yes, I do. Essentially, the idea is that you proof that something, say $q$, implies something that is, say $p$, and something that isn't, say $neg p$. Hence, $q implies (p wedge neg p)$ Which is absurd: something can't both be and not be at the same time--unless, of course, we are discussing quantum mechanics.
– Mack
Feb 14 '13 at 13:37
Okay, now define your $p$ and $q$.
– Joe Z.
Feb 14 '13 at 13:39
Well, I know that $p$ would definitely be that "$i$ is irrational"; however, I am not sure exactly what $q$ would be. Would $q$ be "the sum of a rational and irrational number?"
– Mack
Feb 14 '13 at 13:42
@Mack q⟹(p∧¬p) isn't saying that something is:both be and not be at the same timeFor instance if p is true then ¬p is false, it's not saying that p=¬p, such would be false which is the same logic here: p∧¬p. Ultimately they're not the same, and they are logical opposites.
– CTS_AE
May 17 '17 at 4:38
add a comment |Â
1
Do you understand proof by contradiction?
– Thomas Andrews
Feb 14 '13 at 13:31
@ThomasAndrews Yes, I do. Essentially, the idea is that you proof that something, say $q$, implies something that is, say $p$, and something that isn't, say $neg p$. Hence, $q implies (p wedge neg p)$ Which is absurd: something can't both be and not be at the same time--unless, of course, we are discussing quantum mechanics.
– Mack
Feb 14 '13 at 13:37
Okay, now define your $p$ and $q$.
– Joe Z.
Feb 14 '13 at 13:39
Well, I know that $p$ would definitely be that "$i$ is irrational"; however, I am not sure exactly what $q$ would be. Would $q$ be "the sum of a rational and irrational number?"
– Mack
Feb 14 '13 at 13:42
@Mack q⟹(p∧¬p) isn't saying that something is:both be and not be at the same timeFor instance if p is true then ¬p is false, it's not saying that p=¬p, such would be false which is the same logic here: p∧¬p. Ultimately they're not the same, and they are logical opposites.
– CTS_AE
May 17 '17 at 4:38
1
1
Do you understand proof by contradiction?
– Thomas Andrews
Feb 14 '13 at 13:31
Do you understand proof by contradiction?
– Thomas Andrews
Feb 14 '13 at 13:31
@ThomasAndrews Yes, I do. Essentially, the idea is that you proof that something, say $q$, implies something that is, say $p$, and something that isn't, say $neg p$. Hence, $q implies (p wedge neg p)$ Which is absurd: something can't both be and not be at the same time--unless, of course, we are discussing quantum mechanics.
– Mack
Feb 14 '13 at 13:37
@ThomasAndrews Yes, I do. Essentially, the idea is that you proof that something, say $q$, implies something that is, say $p$, and something that isn't, say $neg p$. Hence, $q implies (p wedge neg p)$ Which is absurd: something can't both be and not be at the same time--unless, of course, we are discussing quantum mechanics.
– Mack
Feb 14 '13 at 13:37
Okay, now define your $p$ and $q$.
– Joe Z.
Feb 14 '13 at 13:39
Okay, now define your $p$ and $q$.
– Joe Z.
Feb 14 '13 at 13:39
Well, I know that $p$ would definitely be that "$i$ is irrational"; however, I am not sure exactly what $q$ would be. Would $q$ be "the sum of a rational and irrational number?"
– Mack
Feb 14 '13 at 13:42
Well, I know that $p$ would definitely be that "$i$ is irrational"; however, I am not sure exactly what $q$ would be. Would $q$ be "the sum of a rational and irrational number?"
– Mack
Feb 14 '13 at 13:42
@Mack q⟹(p∧¬p) isn't saying that something is:
both be and not be at the same time For instance if p is true then ¬p is false, it's not saying that p=¬p, such would be false which is the same logic here: p∧¬p. Ultimately they're not the same, and they are logical opposites.– CTS_AE
May 17 '17 at 4:38
@Mack q⟹(p∧¬p) isn't saying that something is:
both be and not be at the same time For instance if p is true then ¬p is false, it's not saying that p=¬p, such would be false which is the same logic here: p∧¬p. Ultimately they're not the same, and they are logical opposites.– CTS_AE
May 17 '17 at 4:38
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
He starts by assuming you can find rational $r$ and irrational $i$ that have rational sum $s$. Using a direct proof that the difference of two rationals is rational, he shows that this assumption leads to a contradiction. Therefore you cannot find a rational and irrational that sum to a rational, so the sum of a rational and irrational is always irrational.
answered Feb 14 '13 at 13:33
Ross Millikan
278k21188354
1
Oh, I see. It's as if we have two cases: a rational + irrational can equal either a rational or irrational; we've ruled out that it can't be rational, therefore, it must be irrational. Is that correct thinking?
– Mack
Feb 14 '13 at 13:39
1
Yes. Basically, the definition of "irrational" is "not rational." So when he shows that a number is irrational and rational, he has his contradiction.
– Thomas Andrews
Feb 14 '13 at 13:46
add a comment |Â
up vote
0
down vote
Actually he proves more than that: he shows that if $r$ is rational and $iinmathbb R$, then the sum $r+i$ is rational if and only if the number $i$ is rational:
$$
rinmathbb Q ~~textand~~ r+iinmathbb Q quadRightarrowquad iinmathbb Q
$$
hence
$$
rinmathbb Q ~~textand~~ inotinmathbb Q quadRightarrowquad r+inotinmathbb Q
$$
answered Feb 14 '13 at 13:33
AndreasT
3,2661224
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
He starts by assuming you can find rational $r$ and irrational $i$ that have rational sum $s$. Using a direct proof that the difference of two rationals is rational, he shows that this assumption leads to a contradiction. Therefore you cannot find a rational and irrational that sum to a rational, so the sum of a rational and irrational is always irrational.
answered Feb 14 '13 at 13:33
Ross Millikan
278k21188354
1
Oh, I see. It's as if we have two cases: a rational + irrational can equal either a rational or irrational; we've ruled out that it can't be rational, therefore, it must be irrational. Is that correct thinking?
– Mack
Feb 14 '13 at 13:39
1
Yes. Basically, the definition of "irrational" is "not rational." So when he shows that a number is irrational and rational, he has his contradiction.
– Thomas Andrews
Feb 14 '13 at 13:46
add a comment |Â
up vote
5
down vote
accepted
He starts by assuming you can find rational $r$ and irrational $i$ that have rational sum $s$. Using a direct proof that the difference of two rationals is rational, he shows that this assumption leads to a contradiction. Therefore you cannot find a rational and irrational that sum to a rational, so the sum of a rational and irrational is always irrational.
answered Feb 14 '13 at 13:33
Ross Millikan
278k21188354
1
Oh, I see. It's as if we have two cases: a rational + irrational can equal either a rational or irrational; we've ruled out that it can't be rational, therefore, it must be irrational. Is that correct thinking?
– Mack
Feb 14 '13 at 13:39
1
Yes. Basically, the definition of "irrational" is "not rational." So when he shows that a number is irrational and rational, he has his contradiction.
– Thomas Andrews
Feb 14 '13 at 13:46
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
He starts by assuming you can find rational $r$ and irrational $i$ that have rational sum $s$. Using a direct proof that the difference of two rationals is rational, he shows that this assumption leads to a contradiction. Therefore you cannot find a rational and irrational that sum to a rational, so the sum of a rational and irrational is always irrational.
answered Feb 14 '13 at 13:33
Ross Millikan
278k21188354
He starts by assuming you can find rational $r$ and irrational $i$ that have rational sum $s$. Using a direct proof that the difference of two rationals is rational, he shows that this assumption leads to a contradiction. Therefore you cannot find a rational and irrational that sum to a rational, so the sum of a rational and irrational is always irrational.
answered Feb 14 '13 at 13:33
Ross Millikan
278k21188354
answered Feb 14 '13 at 13:33
Ross Millikan
278k21188354
answered Feb 14 '13 at 13:33
Ross Millikan
278k21188354
answered Feb 14 '13 at 13:33
Ross Millikan
278k21188354
278k21188354
1
Oh, I see. It's as if we have two cases: a rational + irrational can equal either a rational or irrational; we've ruled out that it can't be rational, therefore, it must be irrational. Is that correct thinking?
– Mack
Feb 14 '13 at 13:39
1
Yes. Basically, the definition of "irrational" is "not rational." So when he shows that a number is irrational and rational, he has his contradiction.
– Thomas Andrews
Feb 14 '13 at 13:46
add a comment |Â
1
Oh, I see. It's as if we have two cases: a rational + irrational can equal either a rational or irrational; we've ruled out that it can't be rational, therefore, it must be irrational. Is that correct thinking?
– Mack
Feb 14 '13 at 13:39
1
Yes. Basically, the definition of "irrational" is "not rational." So when he shows that a number is irrational and rational, he has his contradiction.
– Thomas Andrews
Feb 14 '13 at 13:46
1
1
Oh, I see. It's as if we have two cases: a rational + irrational can equal either a rational or irrational; we've ruled out that it can't be rational, therefore, it must be irrational. Is that correct thinking?
– Mack
Feb 14 '13 at 13:39
Oh, I see. It's as if we have two cases: a rational + irrational can equal either a rational or irrational; we've ruled out that it can't be rational, therefore, it must be irrational. Is that correct thinking?
– Mack
Feb 14 '13 at 13:39
1
1
Yes. Basically, the definition of "irrational" is "not rational." So when he shows that a number is irrational and rational, he has his contradiction.
– Thomas Andrews
Feb 14 '13 at 13:46
Yes. Basically, the definition of "irrational" is "not rational." So when he shows that a number is irrational and rational, he has his contradiction.
– Thomas Andrews
Feb 14 '13 at 13:46
add a comment |Â
up vote
0
down vote
Actually he proves more than that: he shows that if $r$ is rational and $iinmathbb R$, then the sum $r+i$ is rational if and only if the number $i$ is rational:
$$
rinmathbb Q ~~textand~~ r+iinmathbb Q quadRightarrowquad iinmathbb Q
$$
hence
$$
rinmathbb Q ~~textand~~ inotinmathbb Q quadRightarrowquad r+inotinmathbb Q
$$
answered Feb 14 '13 at 13:33
AndreasT
3,2661224
add a comment |Â
up vote
0
down vote
Actually he proves more than that: he shows that if $r$ is rational and $iinmathbb R$, then the sum $r+i$ is rational if and only if the number $i$ is rational:
$$
rinmathbb Q ~~textand~~ r+iinmathbb Q quadRightarrowquad iinmathbb Q
$$
hence
$$
rinmathbb Q ~~textand~~ inotinmathbb Q quadRightarrowquad r+inotinmathbb Q
$$
answered Feb 14 '13 at 13:33
AndreasT
3,2661224
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Actually he proves more than that: he shows that if $r$ is rational and $iinmathbb R$, then the sum $r+i$ is rational if and only if the number $i$ is rational:
$$
rinmathbb Q ~~textand~~ r+iinmathbb Q quadRightarrowquad iinmathbb Q
$$
hence
$$
rinmathbb Q ~~textand~~ inotinmathbb Q quadRightarrowquad r+inotinmathbb Q
$$
answered Feb 14 '13 at 13:33
AndreasT
3,2661224
Actually he proves more than that: he shows that if $r$ is rational and $iinmathbb R$, then the sum $r+i$ is rational if and only if the number $i$ is rational:
$$
rinmathbb Q ~~textand~~ r+iinmathbb Q quadRightarrowquad iinmathbb Q
$$
hence
$$
rinmathbb Q ~~textand~~ inotinmathbb Q quadRightarrowquad r+inotinmathbb Q
$$
answered Feb 14 '13 at 13:33
AndreasT
3,2661224
answered Feb 14 '13 at 13:33
AndreasT
3,2661224
answered Feb 14 '13 at 13:33
AndreasT
3,2661224
answered Feb 14 '13 at 13:33
AndreasT
3,2661224
3,2661224
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f303963%2fproof-by-contradiction-with-rational-and-irrational-numbers%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Do you understand proof by contradiction?
– Thomas Andrews
Feb 14 '13 at 13:31
@ThomasAndrews Yes, I do. Essentially, the idea is that you proof that something, say $q$, implies something that is, say $p$, and something that isn't, say $neg p$. Hence, $q implies (p wedge neg p)$ Which is absurd: something can't both be and not be at the same time--unless, of course, we are discussing quantum mechanics.
– Mack
Feb 14 '13 at 13:37
Okay, now define your $p$ and $q$.
– Joe Z.
Feb 14 '13 at 13:39
Well, I know that $p$ would definitely be that "$i$ is irrational"; however, I am not sure exactly what $q$ would be. Would $q$ be "the sum of a rational and irrational number?"
– Mack
Feb 14 '13 at 13:42
@Mack q⟹(p∧¬p) isn't saying that something is:
both be and not be at the same timeFor instance if p is true then ¬p is false, it's not saying that p=¬p, such would be false which is the same logic here: p∧¬p. Ultimately they're not the same, and they are logical opposites.– CTS_AE
May 17 '17 at 4:38