Inequlities defination.
Please, give the "accurate" defination of the following inequalities:
1.$ a \le b $
2.$ a\ge b$
3.$a \le b$
4.$a<x<b$
5.$a\le x<b$
6.$a \le x \le b$
7.$a< x\le b$
and, how u prove that, if
i)$a=b$
ii)$a<b$
iii)$a>b$
two of upper three is not true then the other one is true?
1.$ a \le b $
2.$ a\ge b$
3.$a \le b$
4.$a<x<b$
5.$a\le x<b$
6.$a \le x \le b$
7.$a< x\le b$
and, how u prove that, if
i)$a=b$
ii)$a<b$
iii)$a>b$
two of upper three is not true then the other one is true?
A man is not finished when he's defeated, he's finished when he quits.
Re: Inequlities defination.
It's important.
Cause,i ask u to take (a,b) such that $a\nless b$ from the set $\{ 1,2,3,4,...14,15 \}$. Hence, this ques is same meaning to take (a,b) such that $a \ge b$ from the set $\{ 1,2,3,4....14,15 \}$.
But, if i ask u to prove that, $9R^2 \ge a^2+b^2+c^2$. This question isn't same meaning to prove $9R^2 \nless a^2+b^2+c^2$
hence R is circumradious and a,b,c are the hand of triangle
whats the matter?
Cause,i ask u to take (a,b) such that $a\nless b$ from the set $\{ 1,2,3,4,...14,15 \}$. Hence, this ques is same meaning to take (a,b) such that $a \ge b$ from the set $\{ 1,2,3,4....14,15 \}$.
But, if i ask u to prove that, $9R^2 \ge a^2+b^2+c^2$. This question isn't same meaning to prove $9R^2 \nless a^2+b^2+c^2$
hence R is circumradious and a,b,c are the hand of triangle
whats the matter?
A man is not finished when he's defeated, he's finished when he quits.
Re: Inequlities defination.
$9R^2 \ge a^2+b^2+c^2$ and $9R^2 \nless a^2+b^2+c^2$ are the same.
I am not sure what you mean by "definition" of the statements.
The second part of your question is answered by the definition of real numbers. By definition, for any real number $a$ and $b$ either $a \leq b$ or $b \leq a$. So, at least one of the three statements you stated has to be true.
I am not sure what you mean by "definition" of the statements.
The second part of your question is answered by the definition of real numbers. By definition, for any real number $a$ and $b$ either $a \leq b$ or $b \leq a$. So, at least one of the three statements you stated has to be true.
Re: Inequlities defination.
No bro.
$9R^2 \ge a^2+b^2+c^2$ is n't equal to $9R^2 \nless a^2+b^2+c^2$.
Because, $9R^2\ge a^2+b^2+c^2$ means that, $9R^2>a^2+b^2+c^2 \,or\,9R^2=a^2+b^2+c^2$
but, if i prove, $9R^2\nless a^2+b^2+c^2$, it also meant that $9R^2>a^2+b^2+c^2$ is true for all a,b,c. It also meant that only $9R^2=a^2+b^2+c^2$ is true for all a,b,c.
$9R^2 \ge a^2+b^2+c^2$ is n't equal to $9R^2 \nless a^2+b^2+c^2$.
Because, $9R^2\ge a^2+b^2+c^2$ means that, $9R^2>a^2+b^2+c^2 \,or\,9R^2=a^2+b^2+c^2$
but, if i prove, $9R^2\nless a^2+b^2+c^2$, it also meant that $9R^2>a^2+b^2+c^2$ is true for all a,b,c. It also meant that only $9R^2=a^2+b^2+c^2$ is true for all a,b,c.
A man is not finished when he's defeated, he's finished when he quits.
Re: Inequlities defination.
No, you are wrong. $A \nless B$ is same as $A \ge B$.
$A \nless B$ means $A$ is not less than $B$ i.e. $A$ is greater than or equal to $B$ i.e. $A \ge B$.
$A \nless B$ means $A$ is not less than $B$ i.e. $A$ is greater than or equal to $B$ i.e. $A \ge B$.
Re: Inequlities defination.
Oh, then what is meant by $a<>b$ ?
A man is not finished when he's defeated, he's finished when he quits.
Re: Inequlities defination.
nothing. I have never seen $a<>b$.
Re: Inequlities defination.
It appeared in one bdmo question
A man is not finished when he's defeated, he's finished when he quits.
Re: Inequlities defination.
What was the question?