Prove that there are infinitely many primes of the form 3n+ 2 for a positive integern. [This is§4.1Problem 1, which comes with hint you might choose to use. You do not read to read beyond thefirst paragraph of§4.1.]

Proof with contradiction

Step-by-step explanation:

Suppose there only finite odd  prime numbers of the form 3n+2, and we call them . Consider the number . We can see that .

N is a odd number greater than 1, so it is a product of prime numbrs. We easily can see that N can be the product only of prime nubmbers of the form 3n+1 (because production of numbers od the form of 3n+1 is also number of the form 3n+1, and N is not of that form). So some prime number p of the form 3k+2 must divide N. We know that p can't be any prime number of the . Finally we have that p is also a prime number and it isn't for any i less or equal to s. So the set of prime numbers of the form 3n+1 can't be finite.

We speak only of primes numbers of form 3n+1 and 3n+2, becaus numbers of the form 3n are not the prime number but 3. And forms for example 3n+4 is 3n+3+1=3(n+1)+1=3k+1. So we have just those two forms if we multiple n with 3.

he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write   -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get

and solving the given expression :

3x(2x-1)

since solving the expression using tiles   and solving the expression algebraically we get the same answer

thus   we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".