Proof by Contradiction - There exists a positive number $x$ such that $x - frac2x <1$ and $x leq 2$.
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I want to prove that there exists a positive number $x$ such that $x - dfrac2x <1$ and $x leq 2$.
Starting a proof by contradiction, I assumed that there is a positive number x such that x - (2/x) > 1 and x <= 2. I multiplied the first inequality by x and got x^2 - x - 2 > 0. I'm confused where to go from here and how to use the fact that x must also be less than or equal to 2.
real-analysis proof-explanation
edited Sep 6 at 4:04
Nosrati
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asked Sep 6 at 3:56
user130272
58111
add a comment |Â
up vote
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I want to prove that there exists a positive number $x$ such that $x - dfrac2x <1$ and $x leq 2$.
Starting a proof by contradiction, I assumed that there is a positive number x such that x - (2/x) > 1 and x <= 2. I multiplied the first inequality by x and got x^2 - x - 2 > 0. I'm confused where to go from here and how to use the fact that x must also be less than or equal to 2.
real-analysis proof-explanation
edited Sep 6 at 4:04
Nosrati
22.5k61748
asked Sep 6 at 3:56
user130272
58111
2
The easiest way to prove existence is to explicitly produce one example. In this case it's not too hard to consider $x = 1$ - it obviously satisfies both conditions. There is nothing left to prove.
– mathguy
Sep 6 at 4:06
For contradiction, you should assume that for all positive $x$ with $x leq 2$, we have $x - frac 2x geq 1$. Then, you should show that this cannot happen. Simplify this to $x^2 -x- 2 geq 0$, which goes to $(x+1)(x-2) geq 0$. Now, the product of two numbers is positive if and only if both have the same sign, so $0 < x leq 2$ and this equation cannot simultaneously have a solution.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 4:08
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó But it is necessary to distinguish the cases $xge 0$ and $x<0$ when we multiply an inequality with $x$.
– Peter
Sep 6 at 7:07
@Peter $0 < x leq 2$ is part of the assumptions, hence the simplification goes through.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 7:55
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó You are right, I overlooked the word "positive"
– Peter
Sep 6 at 7:56
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to prove that there exists a positive number $x$ such that $x - dfrac2x <1$ and $x leq 2$.
Starting a proof by contradiction, I assumed that there is a positive number x such that x - (2/x) > 1 and x <= 2. I multiplied the first inequality by x and got x^2 - x - 2 > 0. I'm confused where to go from here and how to use the fact that x must also be less than or equal to 2.
real-analysis proof-explanation
edited Sep 6 at 4:04
Nosrati
22.5k61748
asked Sep 6 at 3:56
user130272
58111
I want to prove that there exists a positive number $x$ such that $x - dfrac2x <1$ and $x leq 2$.
Starting a proof by contradiction, I assumed that there is a positive number x such that x - (2/x) > 1 and x <= 2. I multiplied the first inequality by x and got x^2 - x - 2 > 0. I'm confused where to go from here and how to use the fact that x must also be less than or equal to 2.
real-analysis proof-explanation
real-analysis proof-explanation
edited Sep 6 at 4:04
Nosrati
22.5k61748
asked Sep 6 at 3:56
user130272
58111
edited Sep 6 at 4:04
Nosrati
22.5k61748
asked Sep 6 at 3:56
user130272
58111
edited Sep 6 at 4:04
Nosrati
22.5k61748
edited Sep 6 at 4:04
Nosrati
22.5k61748
edited Sep 6 at 4:04
Nosrati
22.5k61748
22.5k61748
asked Sep 6 at 3:56
user130272
58111
asked Sep 6 at 3:56
user130272
58111
asked Sep 6 at 3:56
user130272
58111
58111
2
The easiest way to prove existence is to explicitly produce one example. In this case it's not too hard to consider $x = 1$ - it obviously satisfies both conditions. There is nothing left to prove.
– mathguy
Sep 6 at 4:06
For contradiction, you should assume that for all positive $x$ with $x leq 2$, we have $x - frac 2x geq 1$. Then, you should show that this cannot happen. Simplify this to $x^2 -x- 2 geq 0$, which goes to $(x+1)(x-2) geq 0$. Now, the product of two numbers is positive if and only if both have the same sign, so $0 < x leq 2$ and this equation cannot simultaneously have a solution.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 4:08
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó But it is necessary to distinguish the cases $xge 0$ and $x<0$ when we multiply an inequality with $x$.
– Peter
Sep 6 at 7:07
@Peter $0 < x leq 2$ is part of the assumptions, hence the simplification goes through.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 7:55
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó You are right, I overlooked the word "positive"
– Peter
Sep 6 at 7:56
add a comment |Â
2
The easiest way to prove existence is to explicitly produce one example. In this case it's not too hard to consider $x = 1$ - it obviously satisfies both conditions. There is nothing left to prove.
– mathguy
Sep 6 at 4:06
For contradiction, you should assume that for all positive $x$ with $x leq 2$, we have $x - frac 2x geq 1$. Then, you should show that this cannot happen. Simplify this to $x^2 -x- 2 geq 0$, which goes to $(x+1)(x-2) geq 0$. Now, the product of two numbers is positive if and only if both have the same sign, so $0 < x leq 2$ and this equation cannot simultaneously have a solution.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 4:08
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó But it is necessary to distinguish the cases $xge 0$ and $x<0$ when we multiply an inequality with $x$.
– Peter
Sep 6 at 7:07
@Peter $0 < x leq 2$ is part of the assumptions, hence the simplification goes through.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 7:55
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó You are right, I overlooked the word "positive"
– Peter
Sep 6 at 7:56
2
2
The easiest way to prove existence is to explicitly produce one example. In this case it's not too hard to consider $x = 1$ - it obviously satisfies both conditions. There is nothing left to prove.
– mathguy
Sep 6 at 4:06
The easiest way to prove existence is to explicitly produce one example. In this case it's not too hard to consider $x = 1$ - it obviously satisfies both conditions. There is nothing left to prove.
– mathguy
Sep 6 at 4:06
For contradiction, you should assume that for all positive $x$ with $x leq 2$, we have $x - frac 2x geq 1$. Then, you should show that this cannot happen. Simplify this to $x^2 -x- 2 geq 0$, which goes to $(x+1)(x-2) geq 0$. Now, the product of two numbers is positive if and only if both have the same sign, so $0 < x leq 2$ and this equation cannot simultaneously have a solution.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 4:08
For contradiction, you should assume that for all positive $x$ with $x leq 2$, we have $x - frac 2x geq 1$. Then, you should show that this cannot happen. Simplify this to $x^2 -x- 2 geq 0$, which goes to $(x+1)(x-2) geq 0$. Now, the product of two numbers is positive if and only if both have the same sign, so $0 < x leq 2$ and this equation cannot simultaneously have a solution.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 4:08
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó But it is necessary to distinguish the cases $xge 0$ and $x<0$ when we multiply an inequality with $x$.
– Peter
Sep 6 at 7:07
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó But it is necessary to distinguish the cases $xge 0$ and $x<0$ when we multiply an inequality with $x$.
– Peter
Sep 6 at 7:07
@Peter $0 < x leq 2$ is part of the assumptions, hence the simplification goes through.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 7:55
@Peter $0 < x leq 2$ is part of the assumptions, hence the simplification goes through.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 7:55
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó You are right, I overlooked the word "positive"
– Peter
Sep 6 at 7:56
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó You are right, I overlooked the word "positive"
– Peter
Sep 6 at 7:56
add a comment |Â
2 Answers
2
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Take ANY positive number such that:
$$x<2tag1$$
This implies:
$$frac2x>frac22$$
$$frac2x>1$$
$$-frac2x<-1tag2$$
Add (1) and (2) and you get:
$$x-frac2x<1$$
Actually all positive numbers less than 2 are solutions to your problem. The only exception is $x=2$.
answered Sep 6 at 12:22
Oldboy
3,3151323
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up vote
0
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You made a error in "I assumed that there is a positive number x such that x - (2/x) > 1 and x <= 2."
To use proof by contradiction, it should be "For all positive x, x - (2/x) > 1 or x <= 2."
Actually the original statement is wrong. There does NOT exist a positive number x such that x - (2/x) > 1 and x <= 2.
answered Sep 6 at 13:07
Albert
13
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Take ANY positive number such that:
$$x<2tag1$$
This implies:
$$frac2x>frac22$$
$$frac2x>1$$
$$-frac2x<-1tag2$$
Add (1) and (2) and you get:
$$x-frac2x<1$$
Actually all positive numbers less than 2 are solutions to your problem. The only exception is $x=2$.
answered Sep 6 at 12:22
Oldboy
3,3151323
add a comment |Â
up vote
1
down vote
Take ANY positive number such that:
$$x<2tag1$$
This implies:
$$frac2x>frac22$$
$$frac2x>1$$
$$-frac2x<-1tag2$$
Add (1) and (2) and you get:
$$x-frac2x<1$$
Actually all positive numbers less than 2 are solutions to your problem. The only exception is $x=2$.
answered Sep 6 at 12:22
Oldboy
3,3151323
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Take ANY positive number such that:
$$x<2tag1$$
This implies:
$$frac2x>frac22$$
$$frac2x>1$$
$$-frac2x<-1tag2$$
Add (1) and (2) and you get:
$$x-frac2x<1$$
Actually all positive numbers less than 2 are solutions to your problem. The only exception is $x=2$.
answered Sep 6 at 12:22
Oldboy
3,3151323
Take ANY positive number such that:
$$x<2tag1$$
This implies:
$$frac2x>frac22$$
$$frac2x>1$$
$$-frac2x<-1tag2$$
Add (1) and (2) and you get:
$$x-frac2x<1$$
Actually all positive numbers less than 2 are solutions to your problem. The only exception is $x=2$.
answered Sep 6 at 12:22
Oldboy
3,3151323
answered Sep 6 at 12:22
Oldboy
3,3151323
answered Sep 6 at 12:22
Oldboy
3,3151323
answered Sep 6 at 12:22
Oldboy
3,3151323
3,3151323
add a comment |Â
add a comment |Â
up vote
0
down vote
You made a error in "I assumed that there is a positive number x such that x - (2/x) > 1 and x <= 2."
To use proof by contradiction, it should be "For all positive x, x - (2/x) > 1 or x <= 2."
Actually the original statement is wrong. There does NOT exist a positive number x such that x - (2/x) > 1 and x <= 2.
answered Sep 6 at 13:07
Albert
13
add a comment |Â
up vote
0
down vote
You made a error in "I assumed that there is a positive number x such that x - (2/x) > 1 and x <= 2."
To use proof by contradiction, it should be "For all positive x, x - (2/x) > 1 or x <= 2."
Actually the original statement is wrong. There does NOT exist a positive number x such that x - (2/x) > 1 and x <= 2.
answered Sep 6 at 13:07
Albert
13
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You made a error in "I assumed that there is a positive number x such that x - (2/x) > 1 and x <= 2."
To use proof by contradiction, it should be "For all positive x, x - (2/x) > 1 or x <= 2."
Actually the original statement is wrong. There does NOT exist a positive number x such that x - (2/x) > 1 and x <= 2.
answered Sep 6 at 13:07
Albert
13
You made a error in "I assumed that there is a positive number x such that x - (2/x) > 1 and x <= 2."
To use proof by contradiction, it should be "For all positive x, x - (2/x) > 1 or x <= 2."
Actually the original statement is wrong. There does NOT exist a positive number x such that x - (2/x) > 1 and x <= 2.
answered Sep 6 at 13:07
Albert
13
edited Sep 6 at 13:13
answered Sep 6 at 13:07
Albert
13
answered Sep 6 at 13:07
Albert
13
answered Sep 6 at 13:07
Albert
13
13
add a comment |Â
add a comment |Â
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2
The easiest way to prove existence is to explicitly produce one example. In this case it's not too hard to consider $x = 1$ - it obviously satisfies both conditions. There is nothing left to prove.
– mathguy
Sep 6 at 4:06
For contradiction, you should assume that for all positive $x$ with $x leq 2$, we have $x - frac 2x geq 1$. Then, you should show that this cannot happen. Simplify this to $x^2 -x- 2 geq 0$, which goes to $(x+1)(x-2) geq 0$. Now, the product of two numbers is positive if and only if both have the same sign, so $0 < x leq 2$ and this equation cannot simultaneously have a solution.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 4:08
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó But it is necessary to distinguish the cases $xge 0$ and $x<0$ when we multiply an inequality with $x$.
– Peter
Sep 6 at 7:07
@Peter $0 < x leq 2$ is part of the assumptions, hence the simplification goes through.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 7:55
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó You are right, I overlooked the word "positive"
– Peter
Sep 6 at 7:56