Thread: equality condition that need reasoning

1. equality condition that need reasoning

I have been following Krizalid's lesson here. http://www.mathhelpforum.com/math-he...es-132202.html

I do not understand this part.

Why is $x-2\leq0$
or why can't $x-2>0$ too?

By substituting x=3 and 4, I know it doesn't fit the inequality.

But can anyone explain using reasoning?

2. Originally Posted by stupidguy
I have been following Krizalid's lesson here. http://www.mathhelpforum.com/math-he...es-132202.html

I do not understand this part.

Why is $x-2\leq0$
or why can't $x-2>0$ too?

By substituting x=3 and 4, I know it doesn't fit the inequality.

But can anyone explain using reasoning?
To anyone else wanting to comment on this thread, it helps to view Krizalid's .pdf here, and see the context on page 4.

LHS = left hand side, RHS = right hand side, used below in reference to the inequality you're trying to solve.

So Krizalid isn't placing a restriction $x-2\le0$ but rather is putting this forward as a candidate solution set. Since the radicand is nonnegative for this entire interval, and because furthermore the RHS is nonpositive for this interval, thus the LHS is greater than or equal to the RHS and the inequality holds. Thus $(-\infty, 2]$ is part of the solution set.

You said x=3 doesn't satisfy the inequality but actually it does.. see the steps that occur after both sides are squared.

3. Originally Posted by undefined
To anyone else wanting to comment on this thread, it helps to view Krizalid's .pdf here, and see the context on page 4.

LHS = left hand side, RHS = right hand side, used below in reference to the inequality you're trying to solve.

So Krizalid isn't placing a restriction $x-2\le0$ but rather is putting this forward as a candidate solution set. Since the radicand is defined for this entire interval, and because the LHS is therefore nonnegative and the RHS nonpositive for this interval, the LHS is greater than or equal to the RHS and the inequality holds. Thus $(-\infty, 2]$ is part of the solution set.

You said x=3 doesn't satisfy the inequality but actually it does.. see the steps that occur after both sides are squared.

yes, 3 can. I forgot that LHS can be equal to RHS. My bad.

Why cant x-2>0 be another candidate solution set?

4. Also, I do not understand what is ' [ ', ' ] ', ' ( ', ' ) '.

Why cant he just adopt structures like '3>x>2', $x-2\leq0$?

5. Originally Posted by stupidguy
yes, 3 can. I forgot that LHS can be equal to RHS. My bad.

Why cant x-2>0 be another candidate solution set?
$x-2\ge0$ was used to get the 2 in [2,13/4]. We know the inequality must fail when x > 13/4 because then the radicand will be negative.
Originally Posted by stupidguy
Also, I do not understand what is ' [ ', ' ] ', ' ( ', ' ) '.

Why cant he just adopt structures like '3>x>2', $x-2\leq0$?
http://zonalandeducation.com/mmts/mi...lNotation.html

Interval notation is useful for writing concisely.

6. Originally Posted by undefined
$x-2\ge0$ was used to get the 2 in [2,13/4]. We know the inequality must fail when x > 13/4 because then the radicand ceases to be defined.

Interval Notation

Interval notation is useful for writing concisely.
Why did he "]-infinity,-3]" ?

7. Originally Posted by stupidguy
Why did he "]-infinity,-3]" ?
The ISO notation

Interval (mathematics) - Wikipedia, the free encyclopedia

A bit confusing to have so much notation, I know.

8. how did he find $[-3,3]\cap[2,\frac{13}{4}]=[2,3]$?

I am using number line....but he din show....what other special methods are available?

9. Originally Posted by undefined
The ISO notation

Interval (mathematics) - Wikipedia, the free encyclopedia

A bit confusing to have so much notation, I know.
What is the difference between '(2,3)' and ']2,3[' ?

10. $]-\infty, 2]$ and $(-\infty, 2]$ can be used to indicate that " $-\infty$ is not in the set.

how did he find $[3, 3]\cap[2,\frac{13}{4}]= [2, 3]$?
He didn't- you've miswritten the first part. He found that $[-\infty, 3]\cap [2, \frac{13}{4}]= [2, 3]$. That is because $13/4= 3\frac{1}{4}$, slightly larger than three. Because " $\cap$", the intersection of two sets or those point in both, the intersection does not contain any numbers less than 2, since they are not in the interval $[2, \frac{13}{4}]$ and does not contain any numbers larger than 3 since they are not in $]-\infty, 3]$.

11. Originally Posted by stupidguy
how did he find $[3,3]\frown[2,\frac{13}{4}]=[2,3]$?

I am using number line....but he din show....what other special methods are available?
You made a typo and missed the negative sign in [-3, 3] ... basically if you think about it enough you'll realize $[a,b] \cap [c,d]$ = [max(a,c), min(b,d)] and if max(a,c) > min(b,d) then you get the empty set.

A LaTeX tip: use \cap instead of \frown

Originally Posted by stupidguy
What is the difference between '(2,3)' and ']2,3[' ?
These are the same thing, just different notation.

12. $]-\infty, 2]$ and $(-\infty, 2]$ can be used to indicate that " $-\infty$ is not in the set.

how did he find $[3, 3]\cap[2,\frac{13}{4}]= [2, 3]$?
He didn't- you've miswritten the first part. He found that $[-\infty, 3]\cap [2, \frac{13}{4}]= [2, 3]$. That is because $13/4= 3\frac{1}{4}$, slightly larger than three. Because " $\cap$", the intersection of two sets or those point in both, the intersection does not contain any numbers less than 2, since they are not in the interval $[2, \frac{13}{4}]$ and does not contain any numbers larger than 3 since they are not in $]-\infty, 3]$.

13. my bad for not adding a minus to 3. nvrtheless, i understand, he did the intersection in his head! I usually draw a number line.

now that i understand the basics.

can case 1 be x-2<0 instead?
then, the solution be $(-\infty,2)$

As for the condition for case 2, $[2,\frac{13}{4}]$, I do not know how he get that, though i know it is related to eg 6.

Someone pls elaborate.

14. basically if you think about it enough you'll realize $[a,b] \cap [c,d]$ = [max(a,c), min(b,d)] and if max(a,c) > min(b,d) then you get the empty set.
I dun understand wat u mean. If this is important, pls elaborate. thx.

15. Originally Posted by stupidguy
my bad for not adding a minus to 3. nvrtheless, i understand, he did the intersection in his head! I usually draw a number line.

now that i understand the basics.

can case 1 be x-2<0 instead?
then, the solution be $(-\infty,2)$
Yes it's possible to divide into cases $x-2<0$ and $x-2\ge0$. Notice that this completely covers the real number line; using $x-2\le0$ and $x-2\ge0$ counts $\,x=2$ twice (which isn't a problem since we're taking a union anyway).

Originally Posted by stupidguy
As for the condition for case 2, $[2,\frac{13}{4}]$, I do not know how he get that, though i know it is related to eg 6.

Someone pls elaborate.
What happens when you solve $13x-4\ge0$?

Originally Posted by stupidguy
I dun understand wat u mean. If this is important, pls elaborate. thx.
$x\in[a,b]\cap[c,d]$ means

$x\ge a\ \text{AND}\ x\le b\ \text{AND}\ x\ge c\ \text{AND}\ x\le d$.

Consider just the part $x\ge a\ \text{AND}\ x\ge c$. Can you see why this is equivalent to $x\ge\max(a,c)$?