MATH SOLVE

3 months ago

Q:
# Enter T or F depending on whether the statement is true or not. (You must enter T or F -- True and False will not work.) If it is true do a proof and if it is false provide a counter-example (Of course the proofs and counter-examples must be written down on paper!). If a is divisible by 3 then a is divisible by 9. The substraction of 2 rational numbers is rational. A sufficient condition for an integer to be divisible by 8 is that it is divisible by 2. A sufficient condition for an integer to be divisible by 6 is that it is divisible by 2. If a is divisible by 9 then a is divisible by 3. The product of 2 consecutives integers is even. Answer the following questions. 30 division 3 = , 30 mod 3 = -26 division 5 = , - 26 mod 5 = 28 division 4 = , 28 mod 4 = -29 division 10 = , -29 mod 10 = 24 division 9 = , 24 mod 9 = -28 division 6 = , -28 mod 6 = 965255471 mod 101 = 630153353 mod 101 =

Accepted Solution

A:

Answer:See belowStep-by-step explanation:If a is divisible by 3 then a is divisible by 9
FALSE
Counter-example
6 is divisible by 3 but not by 9
The subtraction of 2 rational numbers is rational.
TRUE
Proof
If a, b are two rational numbers
[tex]\large a=\frac{p}{q}\;b=\frac{r}{s}[/tex]
for some integers p, q, r, s.
Then
[tex]\large a-b=\frac{p}{q}-\frac{r}{s}=\frac{ps-qr}{qs}[/tex]
since ps-qr and qs are integers, a-b is rational
A sufficient condition for an integer to be divisible by 8 is that it is divisible by 2
FALSE
Counter-example
4 is divisible by 2 but not by 8
A sufficient condition for an integer to be divisible by 6 is that it is divisible by 2
FALSE
Counter-example
4 is divisible by 2 but not by 6
If a is divisible by 9 then a is divisible by 3.
TRUE
Proof
If a is divisible by 9, then a = 9k for some integer k, but 9=3*3, so a = 3*(3k).
Since 3k is integer, a is divisible also by 3.
The product of 2 consecutive integers is even.
TRUE
Proof
Let p, q be two consecutive integers, then either p is even or odd.
Suppose first p is even. Then
p = 2n and q = 2n+1, so p*q=2n(2n+1)=2n*2n+2n=2(n*2n+n)
since (n*2n+n) is integer p*q is even.
Suppose now p is odd
p = 2n+1 q = 2n+2, then
p*q=(2n+1)(2n+2)=2n*2n+2*2n+2n+2=2(n*2n+2n+n+1)
since (n*2n+2n+n+1) is integer p*q is also even.
Answer the following questions:30 division 3 = 10
30 mod 3 = 0 (the remainder when dividing 30 by 3)
-26 division 5 =-5 (plus remainder -1)
-26 mod 5 = -1 (the remainder when dividing -26 by 5)
28 division 4 =7
28 mod 4 =0
-29 division 10 = -2 (plus remainder -9)
-29 mod 10 = -9
24 division 9 =2 (plus remainder 6)
24 mod 9 =6
-28 division 6= -4 (plus remainder -4)
-28 mod 6 = -4
965255471 mod 101 = 87 (the remainder when dividing 965255471 by 101)
630153353 mod 101 = 11 (the remainder when dividing 630153353 by 101)