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natta225 [31]
3 years ago
14

Prove that there are infinitely many primes of the form 4k + 3, where k is a non-negative integer. [Hint: Suppose that there are

only finitely many such primes q1, q2, . . . , qn, and consider the number 4q1q2 · · · qn − 1.]
Mathematics
2 answers:
Mrac [35]3 years ago
7 0

Answer:

From the explanation below, the number of primes of the form 4k+3 cannot be finite and if that be the case, the opposite is true that there are infinitely many primes of the form (4k +3)

Step-by-step explanation:

Let q1,q2,…,qn be odd primes of the form 4k+3.

We can write their products as P= (q1xq2....... qr) for some r integer; (4q1+3)x(4q2+3)x..... (4qr + 3)

Let's consider the number N, where

N=4q1q2…qn-1.

It is clear that none of the qi divides N, and that 4 does not divide N.

Since N is odd and greater than 1, it is a product of one or more odd primes.

Now, we'll show that at least one of these primes is of the form 4k+3.

The prime divisors of N cannot be all of the shape 4k+1 because the product of any number of not necessarily distinct primes of the form 4k+1 is itself of the form 4k+1.

But N is not of the form 4k+1. So some prime p of the form 4k+3 divides N.

We have already seen that p cannot be one of q,…,qn.

Thus, it follows that given any collection {q1,…,qn} of primes of the form 4k+3, there is a prime p of the same form which is not in the collection.

Thus, the number of primes of the form 4k+3 cannot be finite and if that be the case, the opposite is true that there are infinitely many primes of the form (4k +3)

Simora [160]3 years ago
6 0

Answer:

The prove is as given below

Step-by-step explanation:

Suppose there are only finitely many primes of the form 4k + 3, say {p1, . . . , pk}. Let P denote their product.

Suppose k is even. Then P ≅ 3^k (mod 4) = 9^k/2 (mod 4) = 1 (mod 4).

ThenP + 2 ≅3 (mod 4), has to have a prime factor of the form 4k + 3. But pₓ≠P + 2 for all 1 ≤ i ≤ k as pₓ| P and pₓ≠2. This is a contradiction.

Suppose k is odd. Then P ≅ 3^k (mod 4) = 9^k/2 (mod 4) = 1 (mod 4).

Then P + 4 ≅3 (mod 4), has to have a prime factor of the form 4k + 3. But pₓ≠P + 4 for all 1 ≤ i ≤ k as pₓ| P and pₓ≠4. This is a contradiction.

So this indicates that there are infinite prime numbers of the form 4k+3.

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Darcy and Harper work for different catering companies. Darcy earns a weekly salary of $982 and an 18% commission on her food
PilotLPTM [1.2K]

Answer:

$2500

Step-by-step explanation:

Given that:

Darcy's weekly earning = $982 plus 18% commission on food sales

Harper's weekly earning = $807 plus 25% commision

amount of food sales will result in Darcy and Harper earning the same amount for the week

Let amount of food sales = f

Darcy's earning = Harper's earning

(982 + 18%*f) = (807 + 25% * f)

(982 + 0.18f) = (807 + 0.25f)

982 + 0.18f = 807 + 0.25f

982 - 807 = 0.25f - 0.18f

175 = 0.07f

f = 175 / 0.07

f = 2500

Hence amount of food sales that will result in both Darcy and Harper earning the same amount = $2500

8 0
3 years ago
67
mina [271]

Answer:

Step-by-step explanation:

1)

A. 2 meters to centimeters

1 m = 100 cm

2 m = 2 \times 100 \\2 m = 200 cm = 2 \times 10^2

B)108 centimeters to meters

100 cm =1 m

1 cm = \frac{1}{100} m \\108 cm = \frac{108}{100}=108 \times 10^{-2} m

C)2.49 meters to centimeters

1 m = 100 cm

2.49 m = 2.49 \times 100 = \frac{249}{100} \times 100 = 249 cm

D)50 centimeters to meters

100 cm = 1m

1 cm = \frac{1}{100} m \\50 cm = \frac{50}{100}= 5 \times 10^{-1} m

E)6.3 meters to centimeters

1 m = 100 cm

6.3 m = 6.3 \times 100=\frac{63}{10} \times 100 = 630  cm

F)7 centimeters to meters

100 cm = 1 m

1 cm = \frac{1}{100} m\\\\7 cm = \frac{7}{100} = 7 \times 10^{-2} m

2)

a)4 meters to millimeters

1 m = 1000 mm

4 m = 4000 = 4 \times 10^3 mm

b)1.7 meters to millimeters

1 m = 1000 mm

1.7 m = 1.7 \times 1000= \frac{17}{10} \times 1000=17 \times 10^2 mm

3 0
3 years ago
Please help a girl out :)
Vilka [71]

Answer:

1/6

Step-by-step explanation:

I don't know how to use it using a number line sorry

-1/3 - (-1/2)

Two minuses give us a plus

so:

-1/3 + 1/2

make the denominators the same and what you multiply at the bottom you multiply at the top (check online about adding fractions if you don't understand)

so:

-2/6 + 3/6 = 1/6

8 0
3 years ago
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The total of,
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 198
 475
+620
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Fudgin [204]
First set up the equation:
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C=(72-32)\times\frac{5}{9}
and solve for C
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3 years ago
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