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.
You can only multiply and add with associative property. are you talking about distributive property or identity property
Answer:
No
Step-by-step explanation:
An extraneous solution is a root of a transformed equation which is not a root of the original equation because it was not included in the domain of the original equation.
Ahmed is solving
for x.
His steps were:
![\begin{aligned}2\sqrt[3]{x-7}&=-8\\ \sqrt[3]{x-7}&=-4\\ \left(\sqrt[3]{x-7}\right)^3&=(-4)^3\\ x-7&=-64 \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D2%5Csqrt%5B3%5D%7Bx-7%7D%26%3D-8%5C%5C%20%20%5Csqrt%5B3%5D%7Bx-7%7D%26%3D-4%5C%5C%20%20%5Cleft%28%5Csqrt%5B3%5D%7Bx-7%7D%5Cright%29%5E3%26%3D%28-4%29%5E3%5C%5C%20%20x-7%26%3D-64%20%5Cend%7Baligned%7D)
Since cube roots <u>do not give two solutions when solved</u>, it is <u>not necessary </u>to check his answers for extraneous solutions.