Mixing
Let x and y be real numbers. Prove that if x≤y+ϵ for every positive real number ϵ, then x≤y. Why do we...
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I understand in the solution that we are finding the contrapositive of $x ≤ y$ which is $x > y$ but why do we set set ϵ= $1over2$(x−y) ? Where does the $1over2$ come in and why do we subtract $y$ from $x$?
discrete-mathematics proof-explanation
asked Jan 22 at 17:38
ZSTZST
11
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add a comment |
$begingroup$
I understand in the solution that we are finding the contrapositive of $x ≤ y$ which is $x > y$ but why do we set set ϵ= $1over2$(x−y) ? Where does the $1over2$ come in and why do we subtract $y$ from $x$?
discrete-mathematics proof-explanation
asked Jan 22 at 17:38
ZSTZST
11
$endgroup$
$begingroup$
Welcome to the site! Usually, math is typeset with the help of mathjax here. I do edit your post so you can have a first look how this works (and you may want to consult the help to get more info).
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– Dirk
Jan 22 at 17:43
add a comment |
$begingroup$
I understand in the solution that we are finding the contrapositive of $x ≤ y$ which is $x > y$ but why do we set set ϵ= $1over2$(x−y) ? Where does the $1over2$ come in and why do we subtract $y$ from $x$?
discrete-mathematics proof-explanation
asked Jan 22 at 17:38
ZSTZST
11
$endgroup$
I understand in the solution that we are finding the contrapositive of $x ≤ y$ which is $x > y$ but why do we set set ϵ= $1over2$(x−y) ? Where does the $1over2$ come in and why do we subtract $y$ from $x$?
discrete-mathematics proof-explanation
discrete-mathematics proof-explanation
asked Jan 22 at 17:38
ZSTZST
11
asked Jan 22 at 17:38
ZSTZST
11
edited Jan 22 at 17:56
ZST
asked Jan 22 at 17:38
ZSTZST
11
asked Jan 22 at 17:38
ZSTZST
11
asked Jan 22 at 17:38
ZSTZST
11
11
$begingroup$
Welcome to the site! Usually, math is typeset with the help of mathjax here. I do edit your post so you can have a first look how this works (and you may want to consult the help to get more info).
$endgroup$
– Dirk
Jan 22 at 17:43
add a comment |
$begingroup$
Welcome to the site! Usually, math is typeset with the help of mathjax here. I do edit your post so you can have a first look how this works (and you may want to consult the help to get more info).
$endgroup$
– Dirk
Jan 22 at 17:43
$begingroup$
Welcome to the site! Usually, math is typeset with the help of mathjax here. I do edit your post so you can have a first look how this works (and you may want to consult the help to get more info).
$endgroup$
– Dirk
Jan 22 at 17:43
$begingroup$
Welcome to the site! Usually, math is typeset with the help of mathjax here. I do edit your post so you can have a first look how this works (and you may want to consult the help to get more info).
$endgroup$
– Dirk
Jan 22 at 17:43
add a comment |
2 Answers
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Well, you assume that $x>y$ and then need to find an $epsilon$ such that $xleq y+epsilon$ does not hold. The given $epsilon$ is such that $y+epsilon$ is right in the middle between $y$ and $x$. (Any other $epsilon = c(x-y)$ with $0<c<1$ would work, too.)
answered Jan 22 at 17:42
DirkDirk
8,8402447
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$begingroup$
Suppose that $x>y$. Then
$$x-y>frac{x-y}{2}>0$$
If we let $epsilon=frac{x-y}{2}$ then we can rewrite this as
$$x-y>epsilon > 0$$
or equivalently,
$$x>y+epsilon>y>0$$
In particular, we found a positive $epsilon$ such that $x>y+epsilon$ i.e. $xleq y+epsilon$ doesn't hold.
answered Jan 22 at 17:44
pwerthpwerth
3,265417
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2 Answers
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2 Answers
2
active
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$begingroup$
Well, you assume that $x>y$ and then need to find an $epsilon$ such that $xleq y+epsilon$ does not hold. The given $epsilon$ is such that $y+epsilon$ is right in the middle between $y$ and $x$. (Any other $epsilon = c(x-y)$ with $0<c<1$ would work, too.)
answered Jan 22 at 17:42
DirkDirk
8,8402447
$endgroup$
add a comment |
$begingroup$
Well, you assume that $x>y$ and then need to find an $epsilon$ such that $xleq y+epsilon$ does not hold. The given $epsilon$ is such that $y+epsilon$ is right in the middle between $y$ and $x$. (Any other $epsilon = c(x-y)$ with $0<c<1$ would work, too.)
answered Jan 22 at 17:42
DirkDirk
8,8402447
$endgroup$
add a comment |
$begingroup$
Well, you assume that $x>y$ and then need to find an $epsilon$ such that $xleq y+epsilon$ does not hold. The given $epsilon$ is such that $y+epsilon$ is right in the middle between $y$ and $x$. (Any other $epsilon = c(x-y)$ with $0<c<1$ would work, too.)
answered Jan 22 at 17:42
DirkDirk
8,8402447
$endgroup$
Well, you assume that $x>y$ and then need to find an $epsilon$ such that $xleq y+epsilon$ does not hold. The given $epsilon$ is such that $y+epsilon$ is right in the middle between $y$ and $x$. (Any other $epsilon = c(x-y)$ with $0<c<1$ would work, too.)
answered Jan 22 at 17:42
DirkDirk
8,8402447
answered Jan 22 at 17:42
DirkDirk
8,8402447
answered Jan 22 at 17:42
DirkDirk
8,8402447
answered Jan 22 at 17:42
DirkDirk
8,8402447
8,8402447
add a comment |
add a comment |
$begingroup$
Suppose that $x>y$. Then
$$x-y>frac{x-y}{2}>0$$
If we let $epsilon=frac{x-y}{2}$ then we can rewrite this as
$$x-y>epsilon > 0$$
or equivalently,
$$x>y+epsilon>y>0$$
In particular, we found a positive $epsilon$ such that $x>y+epsilon$ i.e. $xleq y+epsilon$ doesn't hold.
answered Jan 22 at 17:44
pwerthpwerth
3,265417
$endgroup$
add a comment |
$begingroup$
Suppose that $x>y$. Then
$$x-y>frac{x-y}{2}>0$$
If we let $epsilon=frac{x-y}{2}$ then we can rewrite this as
$$x-y>epsilon > 0$$
or equivalently,
$$x>y+epsilon>y>0$$
In particular, we found a positive $epsilon$ such that $x>y+epsilon$ i.e. $xleq y+epsilon$ doesn't hold.
answered Jan 22 at 17:44
pwerthpwerth
3,265417
$endgroup$
add a comment |
$begingroup$
Suppose that $x>y$. Then
$$x-y>frac{x-y}{2}>0$$
If we let $epsilon=frac{x-y}{2}$ then we can rewrite this as
$$x-y>epsilon > 0$$
or equivalently,
$$x>y+epsilon>y>0$$
In particular, we found a positive $epsilon$ such that $x>y+epsilon$ i.e. $xleq y+epsilon$ doesn't hold.
answered Jan 22 at 17:44
pwerthpwerth
3,265417
$endgroup$
Suppose that $x>y$. Then
$$x-y>frac{x-y}{2}>0$$
If we let $epsilon=frac{x-y}{2}$ then we can rewrite this as
$$x-y>epsilon > 0$$
or equivalently,
$$x>y+epsilon>y>0$$
In particular, we found a positive $epsilon$ such that $x>y+epsilon$ i.e. $xleq y+epsilon$ doesn't hold.
answered Jan 22 at 17:44
pwerthpwerth
3,265417
answered Jan 22 at 17:44
pwerthpwerth
3,265417
answered Jan 22 at 17:44
pwerthpwerth
3,265417
answered Jan 22 at 17:44
pwerthpwerth
3,265417
3,265417
add a comment |
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$begingroup$
Welcome to the site! Usually, math is typeset with the help of mathjax here. I do edit your post so you can have a first look how this works (and you may want to consult the help to get more info).
$endgroup$
– Dirk
Jan 22 at 17:43