Which of the following properly uses a polynomial identity to detect if 2^2k−1 is prime?
The proper use of a polynomial identity to detect if 2^2k−1 is prime is:
If 2^2k−1 can be factored as (2^k + 1)(2^k − 1), then it is not prime.
This is known as the difference of squares identity, where a^2 − b^2 = (a + b)(a − b). Therefore, if 2^2k−1 can be expressed as a difference of squares, it is not prime.
Identify the descending numerical order of these fractions and decimals: 0.733, 0.923, 1 2/5, 17/20, 15/18.(1 point)
The descending numerical order of the given fractions and decimals is:
1. 1 2/5
2. 17/20
3. 15/18
4. 0.923
5. 0.733
Calculate the decimal number for each fraction and identify the ascending numerical order of the decimal numbers:
5/8, 1/4, 3/8, 2/3, 6/7
(1 point)
To calculate the decimal number for each fraction, divide the numerator by the denominator:
5/8 = 0.625
1/4 = 0.25
3/8 = 0.375
2/3 = 0.666666... (rounded to 0.67)
6/7 = 0.857142857... (rounded to 0.86)
The ascending numerical order of the decimal numbers is:
1. 0.25
2. 0.375
3. 0.625
4. 0.67
5. 0.86
that's wrong and do it as fractions
Apologies for the confusion. Let's calculate the decimal number for each fraction by dividing the numerator by the denominator:
5/8 = 0.625
1/4 = 0.25
3/8 = 0.375
2/3 = 0.666666...
6/7 = 0.857142857...
Now let's identify the ascending numerical order of the fractions:
1. 1/4
2. 3/8
3. 5/8
4. 2/3
5. 6/7
wrong
Apologies for the incorrect response. Let's calculate the decimal numbers for each fraction:
5/8 = 0.625
1/4 = 0.25
3/8 = 0.375
2/3 ≈ 0.667
6/7 ≈ 0.857
Now let's identify the ascending numerical order of the decimal numbers:
1. 0.25 (1/4)
2. 0.375 (3/8)
3. 0.625 (5/8)
4. 0.667 (2/3)
5. 0.857 (6/7)
To properly use a polynomial identity to detect if 2^2k-1 is prime, we can utilize Fermat's Little Theorem. This theorem states that if p is a prime number, then for any integer a, a^(p-1) is congruent to 1 modulo p.
In the case of 2^2k-1, we can rewrite it as (2^k)^2 - 1. Now, if we let a = 2^k and p = 2^2k-1, then we can apply Fermat's Little Theorem. According to the theorem, if p is prime, then a^(p-1) will be congruent to 1 modulo p.
So, in this case, for 2^2k-1 to be prime, 2^k raised to the power of (2^2k-2) should be congruent to 1 modulo 2^2k-1.
In other words, we need to check if (2^k)^(2^2k-2) is congruent to 1 modulo (2^2k-1).
To check this congruence, you can use modular exponentiation algorithms like repeated squaring or modular exponentiation by binary exponentiation. These algorithms efficiently calculate exponentiation modulo a number.
If the calculated result is indeed congruent to 1 modulo 2^2k-1, then it suggests that 2^2k-1 might be prime. However, it's important to note that this method is not foolproof and can generate potential pseudoprimes. Therefore, further tests like the Lucas-Lehmer test or Miller-Rabin test should be performed to confirm the primality of the number.