Discrete Maths

[FallonXay]

The idea is that in general (i.e. not always), a decimal number that is rational has one part that is terminating, and a part after it that's recurring.

E.g. \(\frac1{12}=0.083333333\dots=0.08\overline{3}\)
The \(0.08\) part is the terminating part
The \(0.00\overline{3}\) is the recurring part

m and r are integers made to distinguish the purpose. Note that this is a proof where you would say let m and n be blah, not 'suppose' m and n be blah.

I believe that the question wants you to take the first r decimal places (or r-1, you can figure that little thing out) to be the terminating part, and the NEXT m-r decimal places to be the recurring part.

Ahh, I see ~ thanks.

Also, how do you do a) and d) of this question? (For d, I'm not really suer about how to use the idea of the if then in the second part - i.e the relevance of 2^(4n+7) )

[RuiAce]

Ahh, I see ~ thanks.

Also, how do you do a) and d) of this question? (For d, I'm not really suer about how to use the idea of the if then in the second part - i.e the relevance of 2^(4n+7) )

Hint: Induction.

Edit: Oh sorry missed part a)

Part a) hint is proof by exhaustion

[FallonXay]

Hint: Induction.

Edit: Oh sorry missed part a)

Part a) hint is proof by exhaustion

Oooooh! okay, thanks

[FallonXay]

Hello 

How would you do this question?

(Thanks again!)

[RuiAce]

Hello 

How would you do this question?

(Thanks again!)

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[FallonXay]

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x = 1 doesn't give 0 though? So would you just do a proof by cases? (Test, 0, 1, 2, 3 mod4)?

Also, for part a, how come you can just substitute it in? Is that just a property of congruency?

Thanks!

[RuiAce]

x = 1 doesn't give 0 though? So would you just do a proof by cases? (Test, 0, 1, 2, 3 mod4)?

Also, for part a, how come you can just substitute it in? Is that just a property of congruency?

Thanks!

My bad, I think I messed up some of my mental arithmetic because it would be 2 mod 4.
In that case, upon exhausting all the cases we see that the converse is true.

Yes, you were taught several properties of congruence back in topic 2.

[FallonXay]

Heyy, just looking over this proof.  In this line:

How come it's 'a' to the power of 'k' to the power of '(m-1)'. Don't we just substitute 'km' into the place where there was originally an 'n' to give k to the power of (km - 1)?

Also, could you clarify this section? I'm a little confused as to what's happening here:

Thanks! 

[RuiAce]

Heyy, just looking over this proof.  In this line:
How come it's 'a' to the power of 'k' to the power of '(m-1)'. Don't we just substitute 'km' into the place where there was originally an 'n' to give k to the power of (km - 1)?

Also, could you clarify this section? I'm a little confused as to what's happening here:

Thanks! 

That last one had typo's in it. One of those k's should've been a 1 and the other an m. Fixed in the original post.
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[FallonXay]

Question 2 of Discrete Exam Working Out

[RuiAce]

Question 2 of Discrete Exam Working Out


(Image removed from quote.)

"Rationality" is just a colloquial adjective with no meaning. The set of rational numbers is what is closed under the four standard operations. "Taking integer powers" is more appropriate than "powers" by itself (note the consequences of raising to the power of pi, for example).

The flow of the proof is fine.

[FallonXay]

"Rationality" is just a colloquial adjective with no meaning. The set of rational numbers is what is closed under the four standard operations. "Taking integer powers" is more appropriate than "powers" by itself (note the consequences of raising to the power of pi, for example).

The flow of the proof is fine.

Ahhk, cheers! Will fix that up

[RuiAce]

It may also be worth noting that closure under + and x is sufficient. This is because if we want to minus, we simply add the negative. Closure under division is also dangerous in that this may imply we may divide by 0, so closure under multiplication is better (just multiply by a multiplicative inverse, e.g. 1/2).

[FallonXay]

It may also be worth noting that closure under + and x is sufficient. This is because if we want to minus, we simply add the negative. Closure under division is also dangerous in that this may imply we may divide by 0, so closure under multiplication is better (just multiply by a multiplicative inverse, e.g. 1/2).

Oooh, very true. But if I only state closure under + and x, should I explicitly state in my proof multiplication by 1/2 instead of multiplication by 2 and addition of -a and - b instead of subtraction of a and b?

[RuiAce]

Oooh, very true. But if I only state closure under + and x, should I explicitly state in my proof multiplication by 1/2 instead of multiplication by 2 and addition of -a and - b instead of subtraction of a and b?

Yes