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.
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?
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?
Then I make it the speed of the person when the distance between the tree and the car isis .
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.
After checking that the operation is well defined, one can proceed to verify thatis commutative and associative.
Moreover, zeroin is the equivalence class and multiplication in is defined as .
Similarlycan 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
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.
Hi. Consider the following problem:
This is my solution:
LHS isfor all x, y.
RHS isfor all y since .
Hence LHS = RHS if and only if
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?
Hi Al-Allo. Your equations are wrong. They should be as follows
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)
p and q are both less than 0
Do you mean more than 0?
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 A and B be finite sets with m and n elements respectively, and suppose :A→B is a mapping. If is injective (one-to-one), then m ≤ n.
Let A and B be finite sets with m and n elements respectively, and suppose :A→B is a mapping. If is surjective (onto), then m ≥ 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.