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Science talk:Math Exam Resources/Courses/MATH220/December 2010

2012-03-31T20:29:09Z

SimonRose: Talk page autocreated when first thread was posted

Thread:Science talk:Math Exam Resources/Courses/MATH220/December 2010/Splitting up of questions

2012-03-31T20:29:08Z

SimonRose: New thread: Splitting up of questions

This exam has a lot of questions with multiple parts. Before I go and create all of the pages, are we sure that each sub-question should have its own page?

Science:MER/Courses/MATH220

2012-03-31T20:26:51Z

SimonRose: /* Goals */

{{MER Course}}{{MER Infobox small}} 

== Course coordinator ==
The course coordinator is [[User:DavidKohler|David]], get in touch with him for more informations!

==== Role of the course coordinator (CC) ====
* The CC keeps the in progress page up to date.
* The CC sets priorities and goals for the coming days (e.g. finish 2011 Exam, or add 2010 Exam).
* Ideally, the CC finds like-minded people who commit to help achieving these goals for this course.
* The CC watches the discussions, once students start to use it.
* Some grad students would prefer to write solutions on paper. While others prefer to just type stuff in, without solving it themselves. We could use the CC's mailbox in Math 125 as an inbox for solutions to exam questions. That requires some organisation to avoid solving problems twice, and some coordination between writers and typers.
* Coordinate with the admins to make exams that are done visible to students (see below).
* Of course, the CC can also delegate any of the above. But he serves as some kind of link between admins and "normal contributors".

== Contributors to this course ==
If you are a contributor willing to spend time working on MATH220 with the coordinator, please watch this page (to get the updates of the discussion page) and add your name to the list!

==== Contributors ====
* [[User:DavidKohler|David]]
* [[User:Konradbe|Bernhard]]
* [[User:SimonRose|Simon]]

== Goals ==
* The April 2005 exam is done!

* The [[Science:Math_Exam_Resources/Courses/MATH220/December_2011|December 2011]] exam is nearly done; one more question needs its solution reviewed.'''

* '''Let's focus on the [[Science:Math_Exam_Resources/Courses/MATH220/December_2010|December 2010]] exam.'''

* If you are an instructor, feel free to show the current wiki to a student in your office hours this coming week, and ask them for their opinion. What do they like, what functionality would they like to see?

== In progress ==

{| class="wikitable sortable"
|+ style="" | List of active flags in MATH220
|-
! style=""| Flag !! style="" | #
! style=""| Request to ...
|- border="0"
|style="background: #FFFFCC;"| [[Science:MER/Courses/MATH220/Flags/CQ|CQ flags]]
|style="text-align:right;background: #FFFFCC;"| {{MER course flag count|flag_name=CQ|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #FFFFCC;"| add to the [[Science:MER/Statements|statement]] of an exam question.
|-
|style="background: #FFFFCC;"| [[Science:MER/Courses/MATH220/Flags/CH|CH flags]]
|style="text-align:right;background: #FFFFCC;"| {{MER course flag count|flag_name=CH|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #FFFFCC;"| add to some [[Science:MER/Hints|hints]] to an exam question.
|-
|style="background: #FFFFCC;"| [[Science:MER/Courses/MATH220/Flags/CS|CS flags]]
|style="text-align:right;background: #FFFFCC;"| {{MER course flag count|flag_name=CS|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #FFFFCC;"| add to some [[Science:MER/Solutions|solution]] of an exam question.
|- 
|style="background: #FFCC99;"| [[Science:MER/Courses/MATH220/Flags/RQ|RQ flags]]
|style="text-align:right;background: #FFCC99;"| {{MER course flag count|flag_name=RQ|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #FFCC99;"| review an exam question [[Science:MER/Statements|statement]].
|-
|style="background: #FFCC99;"| [[Science:MER/Courses/MATH220/Flags/RH|RH flags]]
|style="text-align:right;background: #FFCC99;"| {{MER course flag count|flag_name=RH|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #FFCC99;"| review [[Science:MER/Hints|hints]].
|-
|style="background: #FFCC99;"| [[Science:MER/Courses/MATH220/Flags/RS|RS flags]]
|style="text-align:right;background: #FFCC99;"| {{MER course flag count|flag_name=RS|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #FFCC99;"| review [[Science:MER/Solutions|solutions]].
|- 
|style="background: #FF9999;"| [[Science:MER/Courses/MATH220/Flags/QBQ|QBQ flags]]
|style="text-align:right;background: #FF9999;"| {{MER course flag count|flag_name=QBQ|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #FF9999;"| improve the [[Science:MER/Statements|statement]] of an exam question reviewed as '''bad'''.
|-
|style="background: #FF9999;"| [[Science:MER/Courses/MATH220/Flags/QBH|QBH flags]]
|style="text-align:right;background: #FF9999;"| {{MER course flag count|flag_name=QBH|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #FF9999;"| improve the [[Science:MER/Hints|hint(s)]] of an exam question reviewed as '''bad'''.
|-
|style="background: #FF9999;"| [[Science:MER/Courses/MATH220/Flags/QBS|QBS flags]]
|style="text-align:right;background: #FF9999;"| {{MER course flag count|flag_name=QBS|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #FF9999;"| improve the [[Science:MER/Solutions|solution(s)]] of an exam question reviewed as '''bad'''.
|- 
|style="background: #CCFFCC;"| [[Science:MER/Courses/MATH220/Flags/QGQ|QGQ flags]]
|style="text-align:right;background: #CCFFCC;"| {{MER course flag count|flag_name=QGQ|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #CCFFCC;"| improve the [[Science:MER/Statements|statement]] of an exam question reviewed as '''good'''.
|-
|style="background: #CCFFCC;"| [[Science:MER/Courses/MATH220/Flags/QGH|QGH flags]]
|style="text-align:right;background: #CCFFCC;"| {{MER course flag count|flag_name=QGH|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #CCFFCC;"| improve the [[Science:MER/Hints|hint(s)]] of an exam question reviewed as '''good'''.
|-
|style="background: #CCFFCC;"| [[Science:MER/Courses/MATH220/Flags/QGS|QGS flags]]
|style="text-align:right;background: #CCFFCC;"| {{MER course flag count|flag_name=QGS|course_name={{#titleparts: {{PAGENAME}}|1|-1}}}}
|style="background: #CCFFCC;"| improve the [[Science:MER/Solutions|solution(s)]] of an exam question reviewed as '''good'''.
|- 
|}

== Most recently updated pages in {{#titleparts:{{PAGENAME}}|1|-1}} ==
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== Accomplishments ==
* The [[Science:Math_Exam_Resources/Courses/MATH220/April_2011|April 2011 Exam]] is now complete!

{{MER Exam Lists}}

To create a new exam page, please use the following box.

{{Create MER Exam page}}

[[Category:MER]]
__NOTOC__

Science:Math Exam Resources/Courses/MATH220/December 2010/Question 03 (c)

2012-03-31T20:23:53Z

SimonRose: Created page with "  <!-- second letter is for ..."

[[Category:MER CQ flag]][[Category:MER CH flag]][[Category:MER CS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH220/December 2010/Question 03 (b)

2012-03-31T20:23:14Z

Science:Math Exam Resources/Courses/MATH220/December 2010/Question 03 (a)

2012-03-31T20:22:58Z

Science:Math Exam Resources/Courses/MATH220/December 2010/Question 02 (b)

2012-03-31T20:22:33Z

Science:Math Exam Resources/Courses/MATH220/December 2010/Question 02 (a)

2012-03-31T20:22:18Z

Science:Math Exam Resources/Courses/MATH220/December 2010/Question 01 (f)

2012-03-31T20:21:53Z

Science:Math Exam Resources/Courses/MATH220/December 2010/Question 01 (e)

2012-03-31T20:21:41Z

Science:Math Exam Resources/Courses/MATH220/December 2010/Question 01 (d)

2012-03-31T20:21:26Z

Science:Math Exam Resources/Courses/MATH220/December 2010/Question 01 (c)

2012-03-31T20:21:10Z

Science:Math Exam Resources/Courses/MATH220/December 2010/Question 01 (b)

2012-03-31T20:20:55Z

Science:Math Exam Resources/Courses/MATH220/December 2010/Question 01 (a)

2012-03-31T20:20:40Z

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 06 (b)

2012-03-31T20:17:42Z

SimonRose: Reviewed.

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 06 (b)/Solution 1

2012-03-31T20:17:15Z

SimonRose:

First, we prove the claim true when <math> n = 1 </math>. In this case,

:<math>
\displaystyle 12^{2n-1} + 11^{n+1} = 12^{2(1)-1} + 11^{1+1} = 12 + 121 = 133
</math>

which is clearly divisible by 133. We assume the claim is true for <math> n=k </math>. For <math> n = k +1 </math>, we have

:<math>
\begin{align}
12^{2(k+1)-1} + 11^{(k+1)+1} &= 12^{2k+1} + 11^{k+2}\\
&= 12^{2}\cdot12^{2k-1} + 11\cdot 11^{k+1}\\
&= 144\cdot12^{2k-1} + 11\cdot 11^{k+1}\\
&= (133+11)12^{2k-1} + 11\cdot 11^{k+1}\\
&= 133\cdot12^{2k-1}+11\cdot12^{2k-1} + 11\cdot 11^{k+1}\\
&= 133\cdot12^{2k-1}+11(12^{2k-1} + 11^{k+1})
\end{align}
</math>

Now, in the last line, the first summand is divisible by 133 and the second summand is also divisible by 133 by the induction hypothesis. Hence we must have that <math> 12^{2(k+1)-1} + 11^{(k+1)+1} </math> is divisible by 133 as required.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 06 (b)/Statement

2012-03-31T20:14:30Z

SimonRose:

Use mathematical induction to prove the following statement:

For all <math>n \in \mathbb{N}, 12^{2n-1} + 11^{n+1}</math> is a multiple of 133.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 05 (b)

2012-03-31T20:12:48Z

SimonRose: Reviewed.

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 05 (b)/Solution 1

2012-03-31T20:12:11Z

SimonRose:

Since <math> A </math> is denumerable, there exists a bijection from <math> A </math> to the set of natural numbers. Let <math> f </math> be this bijection, that is

:<math> f : A \to \mathbb{N} </math>

Let <math>a = f^{-1}(1)</math>. Then define a function <math> g </math> from <math> B := A\backslash\{a\} </math> (a proper subset of <math> A </math>) to <math> \mathbb{N} </math> via

:<math>
\begin{align}
g : B &\to \mathbb{N} \qquad
f^{-1}(n) &\mapsto n-1
\end{align}
</math>

for each <math> n \geq 2 </math>. Notice that this is well defined since <math> f </math> is a bijection between <math> A </math> and the set of natural numbers and so each element of <math> B </math> can be expressed in the form <math> f^{-1}(n) </math> for some <math> n \geq 2 </math>.

We claim that this <math> g </math> is bijective.

To show surjectivity, notice that for each <math> m \in \mathbb{N} </math>, we can take <math> f^{-1}(m+1) \in B </math> and see that <math> g(f^{-1}(m+1)) = m </math>.

For injectivity, suppose that <math> g(b) = g(b') </math> for <math> b,b' \in B </math>. The above comments shows that we can write these elements in the form <math> b = f^{-1}(k)</math> and <math> b = f^{-1}(l)</math> for natural numbers <math> k,l</math>. Hence, we have that

:<math> k-1 = g(f^{-1}(k)) = g(b) = g(b') = g(f^{-1}(l)) = l-1 </math>

and so <math> k=l </math>. By definition, we must have that <math> b=b' </math> and so <math> g</math> is injective completing the proof.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 05 (b)/Solution 1

2012-03-31T20:11:21Z

SimonRose:

Since <math> A </math> is denumerable, there exists a bijection from <math> A </math> to the set of natural numbers. Let <math> f </math> be this bijection, that is

:<math> f : A \to \mathbb{N} </math>

Let <math>a = f^{-1}(1)</math>. Then define a function <math> g </math> from <math> B := A\backslash\{a\} </math> (a proper subset of <math> A </math>) to <math> \mathbb{N} </math> via

:<math>
\begin{align}
g : B &\to \mathbb{N} \qquad
f^{-1}(n) &\mapsto n-1
\end{align}
</math>

for each <math> n \geq 2 </math>. Notice that this is well defined since <math> f </math> is a bijection between <math> A </math> and the set of natural numbers and so each element of <math> B </math> can be expressed in the form <math> f^{-1}(n) </math> for each <math> n \geq 2 </math>.

We claim that this <math> g </math> is bijective.

To show surjectivity, notice that for each <math> m \in \mathbb{N} </math>, we can take <math> f^{-1}(m+1) \in B </math> and see that <math> g(f^{-1}(m+1)) = m </math>.

For injectivity, suppose that <math> g(b) = g(b') </math> for <math> b,b' \in B </math>. The above comments shows that we can write these elements in the form <math> b = f^{-1}(k)</math> and <math> b = f^{-1}(l)</math> for natural numbers <math> k,l</math>. Hence, we have that

:<math> k-1 = g(f^{-1}(k)) = g(b) = g(b') = g(f^{-1}(l)) = l-1 </math>

and so <math> k=l </math>. By definition, we must have that <math> b=b' </math> and so <math> g</math> is injective completing the proof.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 05 (a)

2012-03-31T20:08:49Z

SimonRose: Reviewed.

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{{MER Question page}}

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 05 (a)/Solution 1

2012-03-31T20:08:21Z

SimonRose:

A set is denumerable if is in bijective correspondence with the natural numbers. That is, it is denumerable if and only if there exists a bijective function <math>f : X \to \mathbb{N}</math>.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 04

2012-03-30T19:30:26Z

SimonRose: Asking for review

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 04/Solution 1

2012-03-30T19:29:55Z

SimonRose: Created page with "We want to show that <math>f(C \cap D) \subseteq f(C) \cap f(D)</math> and that <math>f(C) \cap f(D) \subseteq f(C \cap D)</math>. Let us start with the first. Suppose that <..."

We want to show that <math>f(C \cap D) \subseteq f(C) \cap f(D)</math> and that <math>f(C) \cap f(D) \subseteq f(C \cap D)</math>. Let us start with the first.

Suppose that <math>y \in f(C \cap D)</math>; that is, there is some <math>x \in C \cap D</math> such that <math>f(x) = y</math>. Since <math>x \in C</math> and <math>x \in D</math>, it follows that <math>y \in f(C)</math> and that <math>y \in f(D)</math>, i.e. that <math>y \in f(C) \cap f(D)</math> as desired.

''Note: This proof in no way uses injectivity; this is a general statement about all functions''.

Now suppose that <math>y \in f(C) \cap f(D)</math>. That is, <math>y \in f(C)</math> and <math>y \in f(D)</math>. Equivalently, there is some <math>x_1 \in C</math> such that <math>f(x_1) = y</math>, and some <math>x_2 \in D</math> such that <math>f(x_2) = y</math>. We claim that <math>x_1 = x_2</math>, and hence <math>x_1 = x_2 \in C \cap D</math>. But this follows immediately from the injectivity of <math>f</math>. It follows now, since <math>f(x_1) = y</math> and <math>x_1 \in C \cap D</math> that <math>y \in f(C\cap D)</math> as claimed.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 04/Hint 1

2012-03-30T19:22:31Z

SimonRose: Created page with "Remember that to show that two sets <math>X, Y</math> are equal, you must show that <math>X \subseteq Y</math> and that <math>Y \subseteq X</math>."

Remember that to show that two sets <math>X, Y</math> are equal, you must show that <math>X \subseteq Y</math> and that <math>Y \subseteq X</math>.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 04/Statement

2012-03-30T19:21:26Z

SimonRose: Created page with "Let <math>f : A \to B</math> be a function. If <math>C \subseteq A</math>, define :<math>f(C) = \{f(x) : x \in C\}</math> Prove that if <math>f</math> is injective, then <ma..."

Let <math>f : A \to B</math> be a function. If <math>C \subseteq A</math>, define

:<math>f(C) = \{f(x) : x \in C\}</math>

Prove that if <math>f</math> is injective, then <math>f(C \cap D) = f(C) \cap f(D)</math> for all sets <math>C, D \subseteq A</math>.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 07 (a)

2012-03-30T19:18:24Z

SimonRose: Reviewed

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 07 (a)/Solution 1

2012-03-30T19:17:59Z

SimonRose:

Multiplying by any non-zero integer (other than 1 or -1) will do the trick. For example, multiplying by 2 gives the map
:<math>\begin{align}
f: \mathbb{Z} &\to \mathbb{Z}\\
n &\mapsto 2n
\end{align}</math>

'''Proof:''' ''recall that this question didn't ask for a proof, just the answer. Since the proofs are of a difficulty that you are expected to be able to handle in this course, we present them here for completeness.''
which is clearly injective since
:<math>
2n = 2m \iff n = m
</math>
and is not surjective since clearly no odd integer will be in the image of the function &fnof;.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 02 (b)

2012-03-30T19:16:53Z

SimonRose: Asking for review

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{{MER Question page}}

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 02 (b)/Solution 1

2012-03-30T19:16:34Z

SimonRose: Created page with "# The statement <math>P(3) \implies Q(4)</math> is "3 is divisible by 4 implies that 17 is divisible by 3". # The converse is <math>Q(4) \implies P(3)</math>; hence in words ..."

# The statement <math>P(3) \implies Q(4)</math> is "3 is divisible by 4 implies that 17 is divisible by 3".
# The converse is <math>Q(4) \implies P(3)</math>; hence in words it is "17 is divisible by 3 implies that 3 is divisible by 4".
# The contrapositive is <math>\neg Q(4) \implies \neg P(3)</math>. In words this is "17 is not divisible by 3 implies that 3 is not divisible by 4".

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 02 (b)/Hint 2

2012-03-30T19:12:57Z

SimonRose: Created page with "The contrapositive of the statement <math>A \implies B</math> is <math>\neg B \implies \neg A</math>."

The contrapositive of the statement <math>A \implies B</math> is <math>\neg B \implies \neg A</math>.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 02 (b)/Hint 1

2012-03-30T19:11:32Z

SimonRose: Created page with "The converse of the statement <math>A \implies B</math> is <math>B \implies A</math>."

The converse of the statement <math>A \implies B</math> is <math>B \implies A</math>.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (b)

2012-03-30T19:10:02Z

SimonRose: Asking for review

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (b)/Solution 1

2012-03-30T19:09:41Z

SimonRose: Created page with "The set <math>X</math> is the ''intersection'' of all of the sets <math>S_\alpha</math>; that is, <math>X</math> is the set of all <math>x</math> such that, for all <math>\alp..."

The set <math>X</math> is the ''intersection'' of all of the sets <math>S_\alpha</math>; that is, <math>X</math> is the set of all <math>x</math> such that, for all <math>\alpha \in I</math> we have <math>x \in S_\alpha</math>.

The set <math>Y</math> is the ''union'' of all of the sets <math>S_\alpha</math>. That is, <math>Y</math> is the set of all <math>y</math> such that there exists an <math>\alpha \in I</math> such that <math>y \in S_\alpha</math>.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 03

2012-03-30T19:05:33Z

SimonRose: Reviewed

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 03/Solution 1

2012-03-30T19:04:45Z

SimonRose:

Using both hints, we can rewrite the statement into equivalent statements as follow
:<math>\begin{align}
\left[(P \Rightarrow Q) \Rightarrow R\right] \lor (\lnot P \lor Q) &\equiv [(\lnot P \lor Q) \Rightarrow R] \lor (\lnot P \lor Q)\\
&\equiv \lnot(\lnot P \lor Q) \lor R \lor (\lnot P) \lor Q\\
&\equiv (P \land \lnot Q) \lor R \lor (\lnot P) \lor Q\\
&\equiv [P \lor R \lor (\lnot P) \lor Q] \land [(\lnot Q) \lor R \lor (\lnot P) \lor Q]
\end{align}</math>
The last equivalent statement is of the type ''A'' AND ''B'' with
:<math>\begin{align}
A &= P \lor R \lor (\lnot P) \lor Q\\
B &= (\lnot Q) \lor R \lor (\lnot P) \lor Q
\end{align}</math>
which both are tautologies since they contain ''P'' OR ''not P'' and respectively ''not Q'' OR ''Q''. So if both ''A'' and ''B'' are tautologies, then so is the conjunction of them into the statement ''A'' AND ''B''.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (e)

2012-03-30T18:25:32Z

SimonRose: Asking for review

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (e)/Solution 1

2012-03-30T18:24:57Z

SimonRose: Created page with "A function <math>f</math> has an inverse if and only if it is ''bijective''. That is, it has an inverse if and only if it is # Injective (or one-to-one) # Surjective (or onto..."

A function <math>f</math> has an inverse if and only if it is ''bijective''. That is, it has an inverse if and only if it is

# Injective (or one-to-one)
# Surjective (or onto).

In such a case, for every <math>y \in B</math>, there is a unique <math>x \in A</math> such that <math>f(x) = y</math>. Using this fact, we define <math>f^{-1}</math> by the rule

:<math>f^{-1}(y) = x</math>

where <math>x</math> is the unique <math>x \in A</math> such that <math>f(x) = y</math>. From the injectivity and surjectivity, this is well defined. Moreover,

:<math>f\big(f^{-1}(y)\big) = f(x) = y</math>

and

:<math>f^{-1}\big(f(x)\big) = f^{-1}(y) = x</math>

and so this really is the inverse of <math>f</math>.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 08

2012-03-30T18:18:20Z

SimonRose: Reviewed

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 08

2012-03-30T18:17:17Z

SimonRose: Reviewed

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (d)

2012-03-28T03:16:57Z

SimonRose: asking for review

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (d)/Solution 1

2012-03-28T03:16:35Z

SimonRose: Created page with "Since the composition <math>\alpha \circ \beta</math> is given by doing <math>\beta</math> first and then <math>\alpha</math>, we must have that :<math>\begin{align} 1 \mapst..."

Since the composition <math>\alpha \circ \beta</math> is given by doing <math>\beta</math> first and then <math>\alpha</math>, we must have that

:<math>\begin{align}
1 \mapsto 2 \mapsto 3 \\
2 \mapsto 1 \mapsto 2 \\
3 \mapsto 3 \mapsto 4 \\
4 \mapsto 4 \mapsto 1
\end{align}</math>

and so we have that <math>\alpha \circ \beta</math> is given by

:<math>\alpha \circ \beta = \begin{pmatrix}1 & 2 & 3 & 4 \\ 3 & 2 & 4 & 1 \end{pmatrix}</math>

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (a)

2012-03-28T03:12:37Z

SimonRose: asking for review

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (a)/Solution 1

2012-03-28T03:12:22Z

SimonRose: Created page with "De Morgan's laws are, for sets <math>X, Y \subseteq Z</math> the following. :<math>(X \cup Y)^c = X^c \cap Y^c</math> and :<math>(X \cap Y^c = X^c \cup Y^c</math>"

De Morgan's laws are, for sets <math>X, Y \subseteq Z</math> the following.

:<math>(X \cup Y)^c = X^c \cap Y^c</math>

and

:<math>(X \cap Y^c = X^c \cup Y^c</math>

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 02 (a)

2012-03-28T03:10:48Z

SimonRose: asking for review

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 02 (a)/Solution 1

2012-03-28T03:10:19Z

SimonRose: Created page with "We note that <math>P(4)</math> is the statement "4 is divisible by 4", which is true. We also note that <math>Q(4)</math> is the statement "17 is divisible by 3" which is fals..."

We note that <math>P(4)</math> is the statement "4 is divisible by 4", which is true. We also note that <math>Q(4)</math> is the statement "17 is divisible by 3" which is false. As such, the implication

:<math>P(4) \implies Q(4)</math>

does not hold, and so the statement

:<math>\forall\, n \in \mathbb{N}, P(n) \implies Q(n)</math>

is false.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (f)/Solution 1

2012-03-28T03:07:01Z

SimonRose:

Strong mathematical induction is a method to prove a statement <math> S(n) </math> is true for all natural numbers <math> n </math>. It is given by the following.

Suppose that for every <math>n \in \mathbb{N}</math>, that we have the implication

:<math>\big(\forall\,k < n, S(k)\big) \implies S(n)</math>

Then for every <math>n \in \mathbb{N}</math>, <math>S(n)</math> is true.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (f)/Solution 1

2012-03-28T03:06:38Z

SimonRose: Fixed this so that it is about strong induction, not induction.

Strong mathematical induction is a method to prove a statement <math> S(n) </math> is true for all natural numbers <math> n </math>. It is given by the following.

Suppose that for every <math>n \in \mathbb{N}</math>, that we have the implication

:<math>\big(\forall\,k < n S(k)\big) \implies S(n)</math>

Then for every <math>n \in \mathbb{N}</math>, <math>S(n)</math> is true.

Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (c)

2012-03-28T02:39:21Z

SimonRose: Reviewed.

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Science:Math Exam Resources/Courses/MATH220/December 2011/Question 02 (b)

2012-03-27T23:45:30Z

SimonRose: Asking for review

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