You are not logged in.
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.
Hence
We want to minimize
subject to the above constraints. Let us then rewrite the above inequalities in terms of and one of and , say .The 1st and 2nd inequalities give
, the 1st and 3rd inequalities give , and the 2nd and 3rd inequalities give . The minimum appears to be 855 however would imply , which does not satisfy the 2nd inequality. So we must instead have . Thus the minimum cost is $870 dollars for 10 of Platter A and 60 of Platter B.Eliminating
givesSo the maximum value of
is .#1. What is the meaning of the following?
Not true. may only be piecewise differentiable, not wholly differentiable.
Let us define a sequence of functions as follows:
where
is chosen so that for continuity. Then if we make by piecing together the functions and their domains, this function satisfies all your given conditions but is not differentiable at any integer value of , where the gradient changes abruptly.When this is substituted into the original equation, the term in
isWe want this to vanish, so any
such that and will do. So we take . HenceThus under the transformation the curve
becomes the hyperbola . Furthermore as the transformationrepresents a clockwise rotation of 45° about the origin followed by an enlargement of
at the origin, the conic section is preserved, i.e. the original curve is indeed a hyperbola.NB: Be careful when using linear transformations on curves: only rotations, reflections and enlargements/contractions by a nonzero factor preserve conic sections. Any other transformation may distort the curve and alter its original nature.
Given that
and
the proof should be straightforward
#6. Rating: Medium
Let A and B be finite sets with m and n elements respectively, and suppose :A→B is a mapping. If is surjective (onto), show that m ≥ n.
3.Sth
South
3.rns
earns
Thanks bobbym. I've hidden my solution.
According to the Old Testament, 31 kings were defeated by Joshua after the children of Israel had crossed the river Jordan (Joshua 12:924).
#5. Rating: Easy
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), show that m ≤ n.
Is there a 0 (zero)?
Using all the numbers:
#4. Rating: Difficult
x and y are real numbers satisfying
Find the maximum and minimum values of
.
anonimnystefy and bobbym:
Perfect! You got it right.
#3 Rating: Medium
Evaluate:
#2. Rating: Medium
Do there exist integers n such that 3[sup]n[/sup]+7[sup]n[/sup] is a perfect square?