Complete solutions to Exercise I(f) 1
Complete Solutions to Exercise I(f)
1.
(a) Proof. We assume are even. By definition (I.12) they can be written as
where and are integers. Consider their addition :
We have is of the form 2(An Integer). By definition (I.12) we conclude that is even.
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(b) Proof. We assume are even. By definition (I.12) they can be written as
where and are integers. Consider their subtraction:
We have is of the form 2(An Integer). By applying definition (I.12) in the direction we conclude that is even.
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(c) Proof. We assume are odd. By definition (I.14) they can be written as
where and are integers. Consider:
We have is of the form 2(An Integer). By applying definition (I.12) in the direction we conclude that is even whenever are odd.
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(d) Proof. Let be an odd number then by (I.14) there is an integer such that . Consider :
We have . By definition (I.14) we conclude that is odd.
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(e) Proof. Let be even. Then by definition (I.12) this can be written as
Let m be odd then by (I.14)this can be written as
Consider :
(I.12) is an even number where m is an integer
(I.14) is an odd number where m is an integer
We have is 2(Integer) +1 therefore by (I.14) the sum is odd.
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(f) Proof. Let be an odd number then by (I.14)there is an integer a such that . Similarly let m be an odd number then there is an integer b such that
. Consider their product :
We have . By applying definition (I.14)in the direction we conclude that the product is odd.
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(g) Proof. Since is even we can write this as
where is an integer
The product is given by
Hence is a multiple of 2 therefore by definition (I.12) we conclude that is even.
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2. (i) is odd is even
Proof. We assume is odd. Since and 1 are odd therefore by proposition (I.15) we have is even.
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(ii) For any integer we have is even
Proof. If is even then by the above proposition in Question 1(g) we have is even. However if is odd then by the above result2(i) we have is even. Hence again by the above proposition in Question 1(g) we have is even.
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3. If is odd then is even
Proof. By the proposition in Question 1(d) we have is odd is odd. Similarly by the proposition in Question 1(f) we have is odd is odd. Hence is odd. Since and 1 are odd therefore by the proposition in Question 1(c) we have is even. This is what was required.
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4. (a) We need to prove .
Proof. Since therefore by definition (I.16) we have .
(b) We need to prove .
Proof. Since therefore by definition (I.16) we have .
(I.12) is an even number where m is an integer
(I.14) is an odd number where m is an integer
(I.15) The sum of two odd numbers is even
(I.16) divides there is an integer such that
(c) We need to prove .
Proof. Since therefore by definition (I.16) we have .
(d) Prove .
Proof. Since therefore by definition (I.16) we have .
(e) Prove .
Proof. Since which is therefore by definition (I.16) we have .
(f) We have to prove
Proof. We have then by proposition (I.18) we have
where and are arbitrary. Putting we have
the required result, .
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(g) Need to prove:
Proof. From there areintegers and such that
and
Multiplying together gives
Since therefore .
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(h) Need to prove: where
Proof. By using definition (I.16) on we know there is an integer, , such that
Dividing through by gives
which implies
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(i) Prove
Proof. From we have integers and such that
and
Multiplying together gives
which is
(I.16) there is an integer such that
(I.18) If and then
By using definition (I.16) in the direction we have which is what was required. ■
5. (a) We need to prove ‘If is odd then ’
Proof. We assume is odd so it can be written as where is an integer. Consider :
We know from the proposition in Question 2(ii) that is even therefore we have
By definition (I.12) we can write where is an integer. Hence we have
Since which means , therefore and this was what we needed to prove.
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(b) We need to prove ‘If is odd then ’
Proof. We assume is odd so it can be written as where is an integer. Consider the first bracket:
Similarly consider second bracket:
Multiplying these together gives
(I.12) is an even number where m is an integer
Let where is an integer. Substituting this into the above we have
We know from the proposition in Question 2(ii) that is even therefore we
can write where is an integer. We have
We have . By definition (I.16) we conclude that .
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6. Show that if the last digit of an integer is even then is even.
Proof. Using the hint we have
The last line says . We assume is even because the given proposition says “if the last digit of an integer is even” and is the last digit. We can write . We have
(Another Integer) therefore
and so by (I.12) we conclude that is even.
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7. Show that if the last digit of an integer is odd then is odd.
Proof. Very similar to the proof of Question 6.
(I.12) is an even number where m is an integer
(I.16) there is an integer such that