Mixing
Clarification on the method used to prove the existence of $sqrt{2}$ in $mathbb{R}$
$begingroup$
Let $A={xin mathbb{R}:x^2leq{2}}$ with $sup A=alpha$.
When proving that the $sqrt{2}$ exists in $mathbb{R}$ using the method of contradiction, do we show that a contradiction arises for the case $alpha^2 lt{2}$ by negating condition $1$ of the definition for least upper bound, and a contradiction for the case $alpha^2 gt{2}$ by negating condition $2$? leading to the conclusion that $alpha^2={2}$ is the only possibility?
Definition:
Let $Asubseteqmathbb{R}$ and let $alphain mathbb{R}$. Define $alpha$ to be the least upper bound for $A$ if and only if the following two conditions are satisfied.
$alpha$ is an upper bound for $A$; and- for all upper bounds $alpha'$ for $A$, we have that $alpha leq{alpha'}$
proof-explanation
$endgroup$
add a comment |
$begingroup$
Let $A={xin mathbb{R}:x^2leq{2}}$ with $sup A=alpha$.
When proving that the $sqrt{2}$ exists in $mathbb{R}$ using the method of contradiction, do we show that a contradiction arises for the case $alpha^2 lt{2}$ by negating condition $1$ of the definition for least upper bound, and a contradiction for the case $alpha^2 gt{2}$ by negating condition $2$? leading to the conclusion that $alpha^2={2}$ is the only possibility?
Definition:
Let $Asubseteqmathbb{R}$ and let $alphain mathbb{R}$. Define $alpha$ to be the least upper bound for $A$ if and only if the following two conditions are satisfied.
$alpha$ is an upper bound for $A$; and- for all upper bounds $alpha'$ for $A$, we have that $alpha leq{alpha'}$
proof-explanation
$endgroup$
$begingroup$
What are condition 1 and condition 2?
$endgroup$
– José Carlos Santos
Jan 23 at 14:10
1
$begingroup$
Apologies, I have edited to include the definition containing the conditions.
$endgroup$
– user503154
Jan 23 at 14:39
1
$begingroup$
Then, yes, you are right.
$endgroup$
– José Carlos Santos
Jan 23 at 14:41
1
$begingroup$
Thank you @JoséCarlosSantos
$endgroup$
– user503154
Jan 23 at 14:42
add a comment |
$begingroup$
Let $A={xin mathbb{R}:x^2leq{2}}$ with $sup A=alpha$.
When proving that the $sqrt{2}$ exists in $mathbb{R}$ using the method of contradiction, do we show that a contradiction arises for the case $alpha^2 lt{2}$ by negating condition $1$ of the definition for least upper bound, and a contradiction for the case $alpha^2 gt{2}$ by negating condition $2$? leading to the conclusion that $alpha^2={2}$ is the only possibility?
Definition:
Let $Asubseteqmathbb{R}$ and let $alphain mathbb{R}$. Define $alpha$ to be the least upper bound for $A$ if and only if the following two conditions are satisfied.
$alpha$ is an upper bound for $A$; and- for all upper bounds $alpha'$ for $A$, we have that $alpha leq{alpha'}$
proof-explanation
$endgroup$
Let $A={xin mathbb{R}:x^2leq{2}}$ with $sup A=alpha$.
When proving that the $sqrt{2}$ exists in $mathbb{R}$ using the method of contradiction, do we show that a contradiction arises for the case $alpha^2 lt{2}$ by negating condition $1$ of the definition for least upper bound, and a contradiction for the case $alpha^2 gt{2}$ by negating condition $2$? leading to the conclusion that $alpha^2={2}$ is the only possibility?
Definition:
Let $Asubseteqmathbb{R}$ and let $alphain mathbb{R}$. Define $alpha$ to be the least upper bound for $A$ if and only if the following two conditions are satisfied.
$alpha$ is an upper bound for $A$; and- for all upper bounds $alpha'$ for $A$, we have that $alpha leq{alpha'}$
proof-explanation
proof-explanation
edited Jan 23 at 14:19
asked Jan 23 at 14:09
user503154
$begingroup$
What are condition 1 and condition 2?
$endgroup$
– José Carlos Santos
Jan 23 at 14:10
1
$begingroup$
Apologies, I have edited to include the definition containing the conditions.
$endgroup$
– user503154
Jan 23 at 14:39
1
$begingroup$
Then, yes, you are right.
$endgroup$
– José Carlos Santos
Jan 23 at 14:41
1
$begingroup$
Thank you @JoséCarlosSantos
$endgroup$
– user503154
Jan 23 at 14:42
add a comment |
$begingroup$
What are condition 1 and condition 2?
$endgroup$
– José Carlos Santos
Jan 23 at 14:10
1
$begingroup$
Apologies, I have edited to include the definition containing the conditions.
$endgroup$
– user503154
Jan 23 at 14:39
1
$begingroup$
Then, yes, you are right.
$endgroup$
– José Carlos Santos
Jan 23 at 14:41
1
$begingroup$
Thank you @JoséCarlosSantos
$endgroup$
– user503154
Jan 23 at 14:42
$begingroup$
What are condition 1 and condition 2?
$endgroup$
– José Carlos Santos
Jan 23 at 14:10
$begingroup$
What are condition 1 and condition 2?
$endgroup$
– José Carlos Santos
Jan 23 at 14:10
1
1
$begingroup$
Apologies, I have edited to include the definition containing the conditions.
$endgroup$
– user503154
Jan 23 at 14:39
$begingroup$
Apologies, I have edited to include the definition containing the conditions.
$endgroup$
– user503154
Jan 23 at 14:39
1
1
$begingroup$
Then, yes, you are right.
$endgroup$
– José Carlos Santos
Jan 23 at 14:41
$begingroup$
Then, yes, you are right.
$endgroup$
– José Carlos Santos
Jan 23 at 14:41
1
1
$begingroup$
Thank you @JoséCarlosSantos
$endgroup$
– user503154
Jan 23 at 14:42
$begingroup$
Thank you @JoséCarlosSantos
$endgroup$
– user503154
Jan 23 at 14:42
add a comment |
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$begingroup$
What are condition 1 and condition 2?
$endgroup$
– José Carlos Santos
Jan 23 at 14:10
1
$begingroup$
Apologies, I have edited to include the definition containing the conditions.
$endgroup$
– user503154
Jan 23 at 14:39
1
$begingroup$
Then, yes, you are right.
$endgroup$
– José Carlos Santos
Jan 23 at 14:41
1
$begingroup$
Thank you @JoséCarlosSantos
$endgroup$
– user503154
Jan 23 at 14:42