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Hint: See the truth table examples in Chapter 2 for guidance.

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Complete the following truth table involving propositions $p$ and $q$.

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$p$	$q$	$p \\Rightarrow q$	$q \\Rightarrow p$	$\\neg p$	$\\neg q$	$( \\neg p ) \\Rightarrow q$	$( \\neg q ) \\Rightarrow ( \\neg p )$	$\\neg ( p \\Rightarrow q )$	$p \\land ( \\neg q )$	$( \\neg p ) \\land ( \\neg q )$
T	T	[[0]]	[[2]]	[[21]]	[[23]]	[[9]]	[[12]]	[[27]]	[[30]]	[[33]]
T	F	[[19]]	[[3]]	[[22]]	[[7]]	[[10]]	[[26]]	[[15]]	[[16]]	[[34]]
F	T	[[1]]	[[20]]	[[5]]	[[24]]	[[11]]	[[13]]	[[28]]	[[31]]	[[35]]
F	F	[[18]]	[[4]]	[[6]]	[[8]]	[[25]]	[[14]]	[[29]]	[[32]]	[[17]]

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Hint for part i: In the truth table of part a, find all the rows which have a \"T\" in the column of $( \\neg q ) \\Rightarrow ( \\neg p )$. For each such row, move along the row until you reach the entry in the column for $p \\Rightarrow q$. Make a note of this truth value. What do you notice each time?

Hint for part ii: In the truth table of part a, find all the rows which have a \"F\" in the column of $( \\neg q ) \\Rightarrow ( \\neg p )$. For each such row, move along the row until you reach the entry in the column for $p \\Rightarrow q$. Make a note of this truth value. What do you notice each time?

Hint for part iii: The definition of \"logically equivalent\" is described as follows. Suppose we have two propositions $A$ and $B$, which we build up from the two propositions $p$ and $q$ (for example, $A$ could be $p \\Rightarrow q$). $A$ and $B$ are logically equivalent if, for every pair of truth values for $p$ and $q$, the truth value for $A$ and the truth value of $B$ are the same.

i) Suppose $( \\neg q ) \\Rightarrow ( \\neg p )$ is true. What is the truth value of $p \\Rightarrow q$? [[0]]

ii) Suppose $( \\neg q ) \\Rightarrow ( \\neg p )$ is false. What is the truth value of $p \\Rightarrow q$? [[1]]

iii) Hence, can we say that \"$( \\neg q ) \\Rightarrow ( \\neg p )$\" and \"$p \\Rightarrow q$\" are logically equivalent? [[2]]

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Yes

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In the truth table of part a, find the row where the truth value of $( \\neg p ) \\Rightarrow q$ is is true and the truth value of $q$ is false. In this row, what is the truth value of $p$?

Suppose $( \\neg p ) \\Rightarrow q$ is is true and $q$ is false. What is the truth value of $p$? [[0]].

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Hint: Remember that the contrapositive of $p \\Rightarrow q$ (i.e. \"if $p$ then $q$\") is $( \\neg q ) \\Rightarrow ( \\neg p )$ (i.e. \"if $\\neg q$ then $\\neg p$\").

Recall also that $\\neg p$ is the negation of $p$. For example, the negation of \"$1<2$\" is \"$1 \\geq 2$\"; the negation of \"$5$ is odd\" is \"$5$ is even\"; the negation of \"I am 18 years old\" is \"I am not 18 years old\".

In this part, and parts e and f, we will prove, by contraposition, the statement \"if $m^2$ is odd, then $m$ is odd\".

What is the contrapositive statement of \"if $m^2$ is odd, then $m$ is odd\"?
[[0]]

If $m^2$ is odd, then $m$ is odd

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If $m^2$ is even, then $m$ is even

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If $m^2$ is even, then $m$ is odd

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If $m$ is odd, then $m^2$ is odd

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If $m$ is odd, then $m^2$ is even

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If $m$ is even, then $m^2$ is odd

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If $m$ is even, then $m^2$ is even

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Hint: See chapter 2 of the lecture notes, where the definition of \"even\" and \"odd\" is used in some examples.

So, it suffices to prove the statement \"if $m$ is even, then $m^2$ is even\".

Since $m$ is even, we know that
[[0]]

There exists an integer $k$ such that $m=4k+2$

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There exists an integer $k$ such that $m=4k$

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There exists an integer $k$ such that $m=2k$

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There exists an integer $k$ such that $m=4k+1$

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We want you to write $m^2$ in the form $2r$ where $r$ is an integer, to show that $m^2$ is even. Given what we deduced about $m$ in part e, $r$ will depend on $k$.

So, we can now write $m^2 = 2($[[0]]$)$, and so $m^2$ is even.
(When entering an expression into the gap above, please use * for multiplication. For example, write 10*k instead of 10k, as the latter may not always be interpreted correctly by the system.)

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Hint for part i: Use the truth table from part a, taking $p$ to be $s$ and $q$ to be $t$.

In this part of the question, and parts h and i, we will prove, by contradiction, the statement \"if $m$ is even, then $m^2$ is even\". It is in fact easier to prove this directly, but we will nonetheless use \"proof by contradiction\", for practice.

It is helpful to use the letter $s$ to represent the proposition \"$m$ is even\", and the letter $t$ to represent the statement \"$m^2$ is even\". So, what we wish to prove is the statement \"$s \\Rightarrow t$\".

We will assume that the negation of the statement is true, and then arrive by valid mathematical reasoning at a contradiction. This will tell us that the negation of the statement is in fact false, and hence the statement itself is true.

i) Now, the negation of \"$s \\Rightarrow t$\" (i.e. \"$\\neg ( s \\Rightarrow t )$\") is logically equivalent to
[[0]]

ii) Hence the negation of \"if $m$ is even, then $m^2$ is even\" is
[[1]]

$s \\land t$

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$s \\lor t$

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$s \\land ( \\neg t )$

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$s \\lor ( \\neg t )$

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$( \\neg s ) \\land ( \\neg t )$

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$( \\neg s ) \\lor ( \\neg t )$

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$( \\neg s ) \\land t$

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$( \\neg s ) \\lor t$

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$m$ is even, and $m^2$ is even

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$m$ is even, and $m^2$ is odd

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$m$ is even, or $m^2$ is even

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$m$ is even, or $m^2$ is odd

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$m$ is odd, and $m^2$ is odd

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So, we now assume that $m$ is even and $m^2$ is odd.

Since $m$ is even, we can write $m=2k$ for some integer $k$. Hence $m^2 = 2($[[0]]$)$, and this shows that $m^2$ is [[1]].

Hence, using all infomation in this part of the question, we can say that $m^2$ is [[2]], which is clearly a [[3]] statement.

Let $u$ represent the statement \"$m^2$ is [[2]]\".

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Hint for part iii: Look at the truth table for part a. In particular, if \"$( \\neg p ) \\Rightarrow q$\" is true and $q$ is false, what does this tell us about $\\neg p$? Now take $p$ to be \"$s \\Rightarrow t$\" and $q$ to be $u$. Given what you wrote for parts i and ii, and what you just learnt from the truth table, what can you say about the statement \"$\\neg ( s \\Rightarrow t )$\"?

i) We showed by valid mathematical reasoning that \"$( \\neg ( s \\Rightarrow t ) )$\" implies \"$u$\". Hence, the statement \"$( \\neg ( s \\Rightarrow t ) ) \\Rightarrow u$\" is [[0]].

ii) Also, we established that $u$ is [[1]].

iii) This implies that the statement \"$\\neg ( s \\Rightarrow t )$\" is a [[2]] statement. (Press \"Show step\" for a hint).

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Week 1 (lecture 2): In this question you will look at truth tables, proof by contraposition, and proof by contradiction.

Please read the following before attempting the question:

If you have not provided an answer to every input gap of a question or part of the question, and you try to submit your answers to the question or part, then you will see the message \"Can not submit answer - check for errors\". In reality your answer has been submitted, but the system is just concerned that you have not submitted an answer to every input gap. For this reason, please ensure that you provide an answer to every input gap in the question or part before submitting. Even if you are unsure of the answer, write down what you think is most likely to be correct; you can always change your answer or retry the question.

As with all questions, there may be parts where you can choose to \"Show steps\". This may give a hint, or it may present sub-parts which will help you to solve that part of the question. Furthermore, remember to always press the \"Show feedback\" button at the end of each part. Sometimes, helpful feedback will be given here, and often it will depend on how correctly you have answered and will link to other parts of the question. Hence, always retry the parts until you obtain full marks, and then look at the feedback again.

Keep in mind that in order to see the feedback for a particular part of a question, you must provide a full (but not necessarily correct) answer to that part. Do not worry though, as you can look at the feedback and then ammend your answer accordingly.

Furthermore, as with all questions, choosing to reveal the answers will only show you the answers which change every time the question is loaded (i.e. answers to randomised questions); the fixed answers will not be revealed.

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This is the question for week 1 of the MA100 course at the LSE. It looks at material from chapter 2.

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