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#26 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

#28 Re: Help Me ! » math » 2013-01-19 13:22:37


Suppose the order is for
of Platter A and
of Platter B. Then we have:



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.

#29 Re: Help Me ! » inequailities » 2013-01-19 07:57:30


Let
and rewrite the inequalities in terms of
and
:


Eliminating

gives

So the maximum value of

is
.

#30 Puzzles and Games » Scientia’s puzzles » 2013-01-18 12:40:44

scientia
Replies: 10

#1. What is the meaning of the following?

#31 Re: Help Me ! » Little derivative problem (prove or give a counterexample) » 2013-01-16 07:22:12

Not true. ƒ 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.

#32 Re: Help Me ! » Represent the following equation as a hyperbola » 2013-01-16 06:50:47


In questions of this sort I tend look for a transformation to get rid of the
term:

When this is substituted into the original equation, the term in

is

We want this to vanish, so any

such that
and
will do. So we take
. Hence

Thus under the transformation the curve

becomes the hyperbola
. Furthermore as the transformation

represents 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.

#33 Re: This is Cool » Odd composite property1 » 2013-01-14 21:45:34

Given that

and

the proof should be straightforward

#34 Re: Exercises » Scientia’s exercises » 2013-01-13 23:58:57

#6. Rating: Medium

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), show that mn.

#35 Re: Puzzles and Games » omitted vowels » 2013-01-13 21:32:30

mathgogocart wrote:

3.Sth

South


mathgogocart wrote:

3.rns

earns

#36 Re: Exercises » Bafflers? » 2013-01-12 22:11:08

Thanks bobbym. I've hidden my solution.

#38 Re: Help Me ! » Wronksian determinant in 2nd order linear DE » 2013-01-12 02:39:21


I see. Well, if
are differentiable, then linear independence implies
for all
. If they are not both differentiable, then it is possible that they are linearly independent yet
. See http://en.wikipedia.org/wiki/Wronskian# … dependence for an example. smile

#39 Re: Help Me ! » Wronksian determinant in 2nd order linear DE » 2013-01-12 00:35:41


Yes, if you can show that
for some
then
for all
.

#40 Re: Puzzles and Games » post prime numbers! » 2013-01-12 00:15:43

According to the Old Testament, 31 kings were defeated by Joshua after the children of Israel had crossed the river Jordan (Joshua 12:9–24).

#41 Re: Exercises » Scientia’s exercises » 2013-01-10 23:48:26


#5. Rating: Easy

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), show that mn.

#45 Re: Exercises » Scientia’s exercises » 2013-01-09 06:00:26


#4. Rating: Difficult

x and y are real numbers satisfying

Find the maximum and minimum values of

.

#47 Re: Exercises » Scientia’s exercises » 2013-01-08 22:53:40

anonimnystefy and bobbym:

Perfect! You got it right. up


#3 Rating: Medium

Evaluate:

#50 Re: Exercises » Scientia’s exercises » 2013-01-08 11:28:46


#2. Rating: Medium

Do there exist integers n such that 3[sup]n[/sup]+7[sup]n[/sup] is a perfect square?

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