Recall that g is called a primitive root modulo p if the powers of g give all nonzero elements of Fp. (a) For which of the following primes is 2 a primitive root modulo p?a. p = 7b. p = 13c. p = 19d. p = 23

p=13 and p=19.

Step-by-step explanation:

Let <2>={2¹,2²,2³,2⁴,2⁵,2⁶,...} be the set of powers of two modulo p (that is, 2^k=2^kmod p in this set). We can verify if 2 is a primitive root modulo p, by computing (by hand if you like) <2>. If <2> is the set of nonzero elements of Fp, then 2 is a primitive root, if not, then 2 is not a primitive root.

In F_7, <2>={2,4,1}≠F_7\{0}

In F_13, <2>={2,4,8,3,6,12,11,9,2,10,7,1}=F_13\{0}

In F_19, <2>={2,4,8,16,13,7,14,9,18,17,15,11,3,6,12,5,10,1}=F_19\{0}

In F_23, <2>={2,4,8,16,9,18,13,3,6,12,1}≠F_23\{0}

x=half

step-by-step explanation:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: