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#1 Re: Puzzles and Games » Scientia’s puzzles » 2013-03-07 08:01:19

@anonimnystefy: Nope.

@Agnishom: Nope.

Try again. smile

#2 Re: Puzzles and Games » Scientia’s puzzles » 2013-03-06 09:00:33

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

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

#3 Re: Puzzles and Games » Scientia’s puzzles » 2013-03-05 07:47:22

Nope. You were actually closer the first time. wink

#4 Re: Puzzles and Games » Scientia’s puzzles » 2013-03-05 04:25:41

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. big_smile

#5 Re: Help Me ! » Quadratic Equation » 2013-03-01 08:18:52

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. sad


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.

#6 Re: Help Me ! » Quadratic Equation » 2013-02-28 06:28:56

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? mad


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? swear

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

#7 Re: Help Me ! » Is there a way to solve this problem? » 2013-02-27 13:17:16


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
.

#9 Re: Help Me ! » Quadratic Equation » 2013-02-27 12:19:00

Let the roots be
and
. Thus

Eliminating

gives

as the required condition.


Test:

have roots in the ratio
.

#10 Re: Help Me ! » Is there a way to solve this problem? » 2013-02-27 12:00:40

You also need to know the diamter of the tree trunk.

#12 Re: Help Me ! » Vector Spaces! » 2013-02-26 02:17:52

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

#14 Re: Help Me ! » no. of divisers » 2013-02-25 04:58:07

The number of positive divisors of

(where the

are distinct primes) is
.

#15 Re: Help Me ! » Properties of addition » 2013-02-18 01:18:00

No,
is constructed by means of either Dedekind cuts or equivalence classes of Cauchy sequences of rationals. The process is very different and much more complicated because you are now constructing an uncountable set from a countable one.

#16 Re: This is Cool » Induction » 2013-02-18 00:41:08

It looks fine to me. However proving stuff about integers is usually much more straightforward: first prove that the property holds for all non-negative integers, then show that it also holds for
where
is a non-negative integer.

#17 Re: Help Me ! » Properties of addition » 2013-02-18 00:08:02

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 as

#18 Re: This is Cool » Induction » 2013-02-17 23:26:02

anonimnystefy wrote:

I think (c) is unneccessary there, and the rest just represents the axiom of induction.

It is necessary. This is the induction law for
, not
.

#19 Re: Help Me ! » Convergence » 2013-02-11 01:39:33

The sequence gif.latex?\sum_na_n is absolutely convergent iff both gif.latex?\sum_{n=0}^{\infty}a_n and gif.latex?\sum_{n=0}^\infty|a_n| converge.

It is conditionally convergent iff gif.latex?\sum_{n=0}^{\infty}a_n converges while gif.latex?\sum_{n=0}^\infty|a_n| diverges.

Examples.

gif.latex?\sum_n\frac{(-1)^n}{2^n} is absolutely convergent. We have gif.latex?\sum_{n=0}^\infty\frac{(-1)^n}{2^n}=1-\frac12+\frac14-\frac18+\cdots=\frac23 and gif.latex?\sum_{n=0}^\infty\left|\frac{(-1)^n}{2^n}\right|=1+\frac12+\frac14+\cdots=2.

gif.latex?\sum_n\frac{(-1)^n}n is conditionally convergent. We have gif.latex?\sum_{n=0}^\infty\frac{(-1)^n}n=1-\frac12+\frac13-\frac14+\cdots=\ln2 while gif.latex?\sum_{n=0}^\infty\left|\frac{(-1)^n}n\right|=1+\frac12+\frac13+\cdots is divergent.

#20 This is Cool » New 17-million-digit Mersenne prime discovered » 2013-02-08 01:35:25

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. faint

http://z13.invisionfree.com/SAKE/index. … wtopic=258

#21 Re: Help Me ! » imaginary numbers » 2013-02-07 23:40:39

So where are the imaginary numbers? eek

#22 Re: Exercises » Is this cool with you? » 2013-01-28 05:37:05

Hi. Consider the following problem:

[align=center]

[/align]

This is my solution:

LHS is

for all x, y.

RHS is

for all y since
.

Hence LHS = RHS if and only if

and

The second equation gives

; substituting into the first equation gives

Hence the solution set is

.

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

#23 Re: Help Me ! » Just to be sure.... (problem) » 2013-01-27 23:06:09

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. (4 times slower than the numbers of pages Shad read.)

X/12: Number of pages Patrick read. (12 times slower than the numbers of pages Shad read.)

Now, I resolve : X/12=4 (pages read by Patrick)

#24 Re: Help Me ! » Hey! Inequality help (conditions) » 2013-01-23 12:52:46

UrgentHelp wrote:

p and q are both less than 0

Do you mean more than 0?


UrgentHelp wrote:

Now how can we manipulate/transform/ get conditions from these two inequalities which must hold: i.e. q+2p>1 and 2p+4p>2??

The two inequalities are the same. Any permitted values of p and q such that
will work.

#25 Re: This is Cool » How to reduce the time checking if two lists contain the same elements » 2013-01-23 01:15:02

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 A and B be finite sets with m and n elements respectively, and suppose ƒ:AB is a mapping. If ƒ is injective (one-to-one), then mn.

• Let A and B be finite sets with m and n elements respectively, and suppose ƒ:AB is a mapping. If ƒ is surjective (onto), then mn.

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 ƒ 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 ƒ is a mapping from A to itself and so (i) if ƒ is injective, A is a subset of B (ii) if ƒ is surjective, B is a subset of A.

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

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