Mixing
Prove by contradiction that a real number that is less than every positive real number cannot be positive
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
"Prove by contradiction that a real number that is less than every positive real number cannot be positive"
[This is what I did, but it is definitely missing something...
Proof:
1)Assume a real number, n, is less than every positive real and cannot be negative
But if n is less than every positive real number then n is less than the smallest positive real number and thus n can only be less than or equal to zero i.e $nle0$
$nle0$ is a contradiction as n cannot be negative
therefore proven by contradiction]
I'm sure that I'm missing something...
proof-verification
asked Jul 21 at 19:38
stochasticmrfox
216
add a comment |Â
up vote
0
down vote
favorite
"Prove by contradiction that a real number that is less than every positive real number cannot be positive"
[This is what I did, but it is definitely missing something...
Proof:
1)Assume a real number, n, is less than every positive real and cannot be negative
But if n is less than every positive real number then n is less than the smallest positive real number and thus n can only be less than or equal to zero i.e $nle0$
$nle0$ is a contradiction as n cannot be negative
therefore proven by contradiction]
I'm sure that I'm missing something...
proof-verification
asked Jul 21 at 19:38
stochasticmrfox
216
1
"then $n$ is less than the smallest positive real number" You seem to be ignoring the fact that there is no smallest positive real number. Zero is not positive. "and thus $n$ can only be less than or equal to zero" This needs proof.
– JMoravitz
Jul 21 at 19:40
To prove this you should start with whatever properties of the order structure of the real numbers you have proved or assumed as axiomatic. Please edit the question to include an attempt of that kind and we may be able to help.
– Ethan Bolker
Jul 21 at 19:43
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
"Prove by contradiction that a real number that is less than every positive real number cannot be positive"
[This is what I did, but it is definitely missing something...
Proof:
1)Assume a real number, n, is less than every positive real and cannot be negative
But if n is less than every positive real number then n is less than the smallest positive real number and thus n can only be less than or equal to zero i.e $nle0$
$nle0$ is a contradiction as n cannot be negative
therefore proven by contradiction]
I'm sure that I'm missing something...
proof-verification
asked Jul 21 at 19:38
stochasticmrfox
216
"Prove by contradiction that a real number that is less than every positive real number cannot be positive"
[This is what I did, but it is definitely missing something...
Proof:
1)Assume a real number, n, is less than every positive real and cannot be negative
But if n is less than every positive real number then n is less than the smallest positive real number and thus n can only be less than or equal to zero i.e $nle0$
$nle0$ is a contradiction as n cannot be negative
therefore proven by contradiction]
I'm sure that I'm missing something...
proof-verification
asked Jul 21 at 19:38
stochasticmrfox
216
asked Jul 21 at 19:38
stochasticmrfox
216
asked Jul 21 at 19:38
stochasticmrfox
216
asked Jul 21 at 19:38
stochasticmrfox
216
216
1
"then $n$ is less than the smallest positive real number" You seem to be ignoring the fact that there is no smallest positive real number. Zero is not positive. "and thus $n$ can only be less than or equal to zero" This needs proof.
– JMoravitz
Jul 21 at 19:40
To prove this you should start with whatever properties of the order structure of the real numbers you have proved or assumed as axiomatic. Please edit the question to include an attempt of that kind and we may be able to help.
– Ethan Bolker
Jul 21 at 19:43
add a comment |Â
1
"then $n$ is less than the smallest positive real number" You seem to be ignoring the fact that there is no smallest positive real number. Zero is not positive. "and thus $n$ can only be less than or equal to zero" This needs proof.
– JMoravitz
Jul 21 at 19:40
To prove this you should start with whatever properties of the order structure of the real numbers you have proved or assumed as axiomatic. Please edit the question to include an attempt of that kind and we may be able to help.
– Ethan Bolker
Jul 21 at 19:43
1
1
"then $n$ is less than the smallest positive real number" You seem to be ignoring the fact that there is no smallest positive real number. Zero is not positive. "and thus $n$ can only be less than or equal to zero" This needs proof.
– JMoravitz
Jul 21 at 19:40
"then $n$ is less than the smallest positive real number" You seem to be ignoring the fact that there is no smallest positive real number. Zero is not positive. "and thus $n$ can only be less than or equal to zero" This needs proof.
– JMoravitz
Jul 21 at 19:40
To prove this you should start with whatever properties of the order structure of the real numbers you have proved or assumed as axiomatic. Please edit the question to include an attempt of that kind and we may be able to help.
– Ethan Bolker
Jul 21 at 19:43
To prove this you should start with whatever properties of the order structure of the real numbers you have proved or assumed as axiomatic. Please edit the question to include an attempt of that kind and we may be able to help.
– Ethan Bolker
Jul 21 at 19:43
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
By contradiction, suppose that the number $x$ less than every positive real number is positive therefore let $y=frac x 2>0$ and we have $y<x$.
answered Jul 21 at 19:43
gimusi
65.4k73583
add a comment |Â
up vote
0
down vote
Assuming that the nonnegative $xinmathbbR$ is less than every positive real number, then there are two possibilities:
1.) $x=0$,
or
2.) $x>0$.
If $x>0$, then $exists yinmathbbR$ such that $y=fracxa$ for some $a>1$. Conventionally $a=2$ is used, and therefore $y<x$ where $y>0$. However, we have assumed that $x$ is smaller than every positive real number, which is in contradiction with the fact that $y<x$ and $yinmathbbR$. Therefore it must be the case that $x=0$.
answered Jul 21 at 19:54
BoisterousLemma
967
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
By contradiction, suppose that the number $x$ less than every positive real number is positive therefore let $y=frac x 2>0$ and we have $y<x$.
answered Jul 21 at 19:43
gimusi
65.4k73583
add a comment |Â
up vote
2
down vote
accepted
By contradiction, suppose that the number $x$ less than every positive real number is positive therefore let $y=frac x 2>0$ and we have $y<x$.
answered Jul 21 at 19:43
gimusi
65.4k73583
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
By contradiction, suppose that the number $x$ less than every positive real number is positive therefore let $y=frac x 2>0$ and we have $y<x$.
answered Jul 21 at 19:43
gimusi
65.4k73583
By contradiction, suppose that the number $x$ less than every positive real number is positive therefore let $y=frac x 2>0$ and we have $y<x$.
answered Jul 21 at 19:43
gimusi
65.4k73583
answered Jul 21 at 19:43
gimusi
65.4k73583
answered Jul 21 at 19:43
gimusi
65.4k73583
answered Jul 21 at 19:43
gimusi
65.4k73583
65.4k73583
add a comment |Â
add a comment |Â
up vote
0
down vote
Assuming that the nonnegative $xinmathbbR$ is less than every positive real number, then there are two possibilities:
1.) $x=0$,
or
2.) $x>0$.
If $x>0$, then $exists yinmathbbR$ such that $y=fracxa$ for some $a>1$. Conventionally $a=2$ is used, and therefore $y<x$ where $y>0$. However, we have assumed that $x$ is smaller than every positive real number, which is in contradiction with the fact that $y<x$ and $yinmathbbR$. Therefore it must be the case that $x=0$.
answered Jul 21 at 19:54
BoisterousLemma
967
add a comment |Â
up vote
0
down vote
Assuming that the nonnegative $xinmathbbR$ is less than every positive real number, then there are two possibilities:
1.) $x=0$,
or
2.) $x>0$.
If $x>0$, then $exists yinmathbbR$ such that $y=fracxa$ for some $a>1$. Conventionally $a=2$ is used, and therefore $y<x$ where $y>0$. However, we have assumed that $x$ is smaller than every positive real number, which is in contradiction with the fact that $y<x$ and $yinmathbbR$. Therefore it must be the case that $x=0$.
answered Jul 21 at 19:54
BoisterousLemma
967
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assuming that the nonnegative $xinmathbbR$ is less than every positive real number, then there are two possibilities:
1.) $x=0$,
or
2.) $x>0$.
If $x>0$, then $exists yinmathbbR$ such that $y=fracxa$ for some $a>1$. Conventionally $a=2$ is used, and therefore $y<x$ where $y>0$. However, we have assumed that $x$ is smaller than every positive real number, which is in contradiction with the fact that $y<x$ and $yinmathbbR$. Therefore it must be the case that $x=0$.
answered Jul 21 at 19:54
BoisterousLemma
967
Assuming that the nonnegative $xinmathbbR$ is less than every positive real number, then there are two possibilities:
1.) $x=0$,
or
2.) $x>0$.
If $x>0$, then $exists yinmathbbR$ such that $y=fracxa$ for some $a>1$. Conventionally $a=2$ is used, and therefore $y<x$ where $y>0$. However, we have assumed that $x$ is smaller than every positive real number, which is in contradiction with the fact that $y<x$ and $yinmathbbR$. Therefore it must be the case that $x=0$.
answered Jul 21 at 19:54
BoisterousLemma
967
answered Jul 21 at 19:54
BoisterousLemma
967
answered Jul 21 at 19:54
BoisterousLemma
967
answered Jul 21 at 19:54
BoisterousLemma
967
967
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%2f2858813%2fprove-by-contradiction-that-a-real-number-that-is-less-than-every-positive-real%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
"then $n$ is less than the smallest positive real number" You seem to be ignoring the fact that there is no smallest positive real number. Zero is not positive. "and thus $n$ can only be less than or equal to zero" This needs proof.
– JMoravitz
Jul 21 at 19:40
To prove this you should start with whatever properties of the order structure of the real numbers you have proved or assumed as axiomatic. Please edit the question to include an attempt of that kind and we may be able to help.
– Ethan Bolker
Jul 21 at 19:43