Reading...
Play button
Play button
Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
9 Cards in this Set
- Front
- Back
|
State the Proof of Divisibility
|
An integer m DIVIDES an integer n : Algebraically shown as (m|n) => (n/m) this is what they imply
An integer 2 divides an integer 4 => (2|4) => (4/2) Only if there exists an integer k so that n=km I.E 4/2 = 2 (n/m=k) ..... 4 = 2 x 2 (n=km) 12/4 = 3 (n/m=k) ..... 12=4 x 3 (n=km) 42/6 = 7 (n/m=k) ..... 42 = 6 x 7 (n=km) WHOLE CONCEPT Switching back and forth from Divison and Multiplication! |
|
|
When you are reading a proof of an implication do the follow.
|
1.Explicitly identify the hypothesis and the conclusion. (no hypothesis is stated then write "No explicit hypothesis")
2. Explicitly identify the core proof technique. 3. Record any preliminary material needed(defns previously proved propositions or techniques) 4. Justify each step with reference to the defn, previously proved proposition or technique used 5. Add missing steps where necessary and justify the these steps with reference to the definitions, previously proved propositions or techniques used. |
|
|
Prove Transitivity of Divisibility
|
1. Since a|b, there exists an integer r so that ra=b
[12/4 = 3] --> [b/a=r] --> 3(4) = 12 [ra=b] 2. Since b|c, there exists an integer s so that sb=c [24/12=2] (c/b=s) --> (2)(12)=24 [sb=c] 3. Substituting ra for b in the previous equation, we get (sr)(a)=c ---> (2)(3)(4) = 24 5. Since sr is an integer, a|c (c/a) |
|
|
State DIC proposition
|
If a,b and c are integer where a|b (b/a) and a|c (b/a) and x and y are any integer then a|(bx+cy) ... (bx+cy/a)
a=24 b=4 c=2 x=6 y=3 --> 24/(4(6)+(2)(3) |
|
|
Prove DIC
|
1. Since a/b there exists an integer r such that b=ra
2. Since a/c there exists an integer s such that c=sa 3. bx+cy = (ra)x + (sa)y = a(rx+sy) 4. Since there exists an integer k so that bx+cy = ka Then a|(bx+cy) |
|
|
Bounds by divisibility state proposition
|
Let a and b be integers.
If a|b and b does not equal 0 Then |a| less than or equal to |b| i.e a=6 b=12 --> 6<12 |
|
|
Prove Bounds by divisibility
|
1.Since a|b, there exists an integer q so that b=qa.
2. Since b not equal 0; q not equal 0 3. But if q does not equal, |q| greater than equal to 1 4. Hence, |b|=|qa| = |q||a| greater than equal to |a| |
|
|
Analysis of Bounds of Divisibility
|
Analysis of Proof: explicitly identify the hypothesis and the conclusion
**** |
|
|
Division Algorithm
|
****
|
