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@anonimnystefy: Nope.

@Agnishom: Nope.

Try again.

If you get the hang of it, you can try the next one.

#2. [1 − cos[sup]2[/sup]()][sup]1/2[/sup]

Nope. You were actually closer the first time.

anonimnystefy wrote:

The plus/minus sign followed by a times sign? Doesn't make much sense...

You are very, **very**, **VERY** close to the answer.

**debjit625**, I'm sorry for my bad-tempered post above. I suppose I had had a bad day and somehow let it spill over onto the forum. It was totally wrong of me and I apologize.

Anyway, "condition" simply means something that has to be fulfilled in order for a given statement to be true. For example, suppose the question was: "What's the condition under which the quadratic equation , , has no real roots?" Then the condition would be: . In order for the statement ", , has no real roots" to be true, the condition has to be fulfilled.

In this particular question, the statement is "

() has roots in the ratio ". In order for this to be true, we must have . This is the condition that has to be fulfilled in order for the given statement to be true.What the bleep is wrong? The question asks for the condition under which certain things hold. I worked out and gave the condition. Just what are you not happy with?

debjit625 wrote:

Ok I think you guys didn't understood what I was asking for...

I want to know what the question is asking for ,what condition ?

Look. You asked a question. I took the time to help you with it. It was late, I could have gone to bed, but I stayed up to help you. Is this the gratitude I get in return?

Maybe what you don't understand is not the question but the English language?

Okay, suppose the car is travelling with velocity , the tree has diameter and is at distance from the road; suppose the hidden person is hugging the tree, and that, looking down from above, the person, car and centre of the cross section of the tree trunk are always in a straight line.

Then I make it the speed of the person when the distance between the tree and the car is

is .Try

No, it doesn't work.

Eliminating

givesas the required condition.

Test:

have roots in the ratio .You also need to know the diamter of the tree trunk.

I think 123ronnie321 is looking for the direct sum, not the union.

Oops, pardon me.

The number of positive divisors of

(where the

are distinct primes) is .anonimnystefy wrote:

It cannot be prpved. It is an axiom of natural numbers and of higher number sets, too.

It depends on how you are defining your sets of numbers. If you are defining them by axioms, then there is nothing to prove since the axioms will include commutativity and associativity of addition. But if you are constructing them from smaller sets of numbers, then commutativity and associativity need to be proved.

For example, here is briefly how is constructed from . Define a relation on by iff . (Hint: Think of as the "difference" .) Then is an equivalence relation and is defined as the set of all equivalence classes under . Let the equivalence class containing be denoted . Then addition in is defined as

After checking that the operation is well defined, one can proceed to verify that

is commutative and associative.Moreover, zero

in is the equivalence class and multiplication in is defined as .Similarly

can be constructed from as equivalence classes of the equivalence relation on defined by iff . Letting the equivalence class containing be denoted addition in is definied asanonimnystefy wrote:

It is necessary. This is the induction law for , not .
I think (c) is unneccessary there, and the rest just represents the axiom of induction.

The sequence is absolutely convergent iff both and converge.

It is conditionally convergent iff converges while diverges.

Examples.

is absolutely convergent. We have and .

is conditionally convergent. We have while is divergent.

**scientia**- Replies: 7

A new prime number has been discovered this week by GIMPS:

It is now the largest known prime, a whopping 17-million-digit Mersenne prime.

So where are the imaginary numbers?

Hi. Consider the following problem:

[align=center]

[/align]This is my solution:

LHS is

for allRHS is

for allHence LHS = RHS if and only if

and

The second equation gives

; substituting into the first equation givesHence the solution set is

.So my question is: Is my proof okay? Is everyone happy with my proof?

Hi Al-Allo. Your equations are wrong. They should be as follows

Al-Allo wrote:

Aaron reads 3 times faster than Patrick and Shad reads 4 times faster than Aaron. If Patrick reads 4 pages in 1 hour, how many pages reads each of his two comrades? I know how to resolve the problem, but I just wanna be sure I resolved it in the right way.

X: Number of pages Shad read

X/4: Number of pages Aaron read. (4timesslowerthan the numbers of pages Shad read.)

X/12: Number of pages Patrick read. (12 timesslowerthan the numbers of pages Shad read.)Now, I resolve :

X/12=4 (pages read by Patrick)

UrgentHelp wrote:

p and q are both less than 0

Do you mean **more** than 0?

UrgentHelp wrote:

The two inequalities are the same. Any permitted values of Now how can we manipulate/transform/ get conditions from these two inequalities which must hold: i.e. q+2p>1 and 2p+4p>2??

Hi Ivar Sand.

Your problem is a consequence of two problems I set in my exercises thread about a mappings: http://www.mathisfunforum.com/viewtopic.php?id=18790:

Let

AandBbe finite sets withmandnelements respectively, and suppose :A→Bis a mapping. If is injective (one-to-one), thenm≤n. Let

AandBbe finite sets withmandnelements respectively, and suppose :A→Bis a mapping. If is surjective (onto), thenm≥n.

Normally to prove that a mapping is bijective, you have to show that it is both injective and surjective. However, if *A* and *B* are finite sets with the same number of elements (i.e. *m* = *n*) then the two statements above imply that will be bijective if we can show that it is EITHER injective OR surjective (so we don't have to waste them showing both).

In particular, if *A* = *B*, we have exactly the same problem as the theorem you are discussing here. This is because in this case is a mapping from *A* to itself and so (i) if is injective, *A* is a subset of *B* (ii) if is surjective, *B* is a subset of *A*.

PS: No-one has had a go at my exercises yet.