. Prove the following: Let a, b, c, n E Z, where n 2 2. If ac = bc (mod n) and gcd(c, n) = 1,then a = b (mod n)
时间: 2024-06-02 08:12:54 浏览: 15
To prove that a = b (mod n), we need to show that n divides (a - b).
Since ac = bc (mod n), we have n divides (ac - bc) = c(a - b).
Since gcd(c, n) = 1, we know that n and c are coprime, which means that n does not divide c.
Therefore, n must divide (a - b), which means that a = b (mod n).
Hence, the statement is proved.
相关问题
12 Prove and disprove (a) Prove or disprove the following claim: 16n = (b) Prove or disprove the following claim: = O(16n)
(a) This claim is incomplete as there is no value or expression given after the equals sign. Therefore, it cannot be proven or disproven without additional information.
(b) This claim is true. To prove it, we need to show that there exists a constant c and an n0 such that 16n ≤ c16n for all n ≥ n0. Let c = 1 and n0 = 1. Then, 16n ≤ 16n for all n ≥ 1, which satisfies the definition of O(16n). Therefore, the claim is proven.
Using the generating function for the Fibonacci numbers, prove the identity f0 +f2 +...+f2n =f2n+1 for any n ≥ 0.
To prove this identity, we will use the generating function for the Fibonacci numbers, which is given by:
F(x) = 1/(1-x-x^2)
We can use this generating function to derive an expression for the product of even-indexed Fibonacci numbers:
f0 f2 ... f2n = F(x^2) = 1/(1-x^2-x^4)...(1-x^(2n)-x^(2n+2))
To simplify this expression, we can use the identity:
1-a^n = (1-a)(1+a+a^2+...+a^(n-1))
Using this identity, we can write:
1-x^(2n+2) = (1-x^2)(1+x^2+x^4+...+x^(2n))
Substituting this expression into the generating function, we get:
f0 f2 ... f2n = 1/(1-x^2-x^4)...(1-x^(2n)(1-x^2)(1+x^2+x^4+...+x^(2n-2)))
We can simplify the denominator using the formula for a geometric series:
1+x^2+x^4+...+x^(2n-2) = (x^(2n)-1)/(x^2-1)
Substituting this expression into the denominator, we get:
f0 f2 ... f2n = 1/(1-x^2-x^4)...(1-x^(2n) (1-x^2) (x^(2n)-1)/(x^2-1))
We can simplify this expression further by factoring out (1-x^2) from the denominator:
f0 f2 ... f2n = (1-x^2)^n / (1-x^2-x^4)...(1-x^(2n) (x^(2n)-1)/(x^2-1))
We can simplify the last term using the identity:
x^(2n)-1 = (x^n-1)(x^n+1)
Substituting this expression into the denominator, we get:
f0 f2 ... f2n = (1-x^2)^n / (1-x^2-x^4)...(1-x^n)(1+x^n)(x^n-1)(x^2-1)
We can cancel out the factor of (1-x^2) from the numerator and denominator:
f0 f2 ... f2n = (1-x^2)^(n-1) / (1-x^4-x^8)...(1-x^n)(1+x^n)(x^n-1)(x^2-1)
Using the identity:
1-x^4-x^8-...-x^(4n) = (1-x^2)(1+x^2+x^4+...+x^(2n))
We can simplify the denominator further:
f0 f2 ... f2n = (1-x^2)^(n-1) / ((1-x^2)(1+x^2+x^4+...+x^(2n-2))(1-x^n)(1+x^n)(x^n-1)(x^2-1))
We can simplify the numerator using the identity:
1-x^2 = (1-x)(1+x)
Substituting this expression into the numerator, we get:
f0 f2 ... f2n = (1-x)^(n-1) (1+x)^(n-1) / ((1-x)(1+x+x^2+...+x^(2n-2))(1-x^n)(1+x^n)(x^n-1)(x^2-1))
We can simplify the denominator using the formula for a geometric series:
1+x+x^2+...+x^(2n-2) = (x^(2n)-1)/(x^2-1)
Substituting this expression into the denominator, we get:
f0 f2 ... f2n = (1-x)^(n-1) (1+x)^(n-1) (x^n+1) / ((1-x)(x^n+1)(x^n-1)(x^2-1))
We can cancel out the factors of (1-x^n) and (x^n+1) from the numerator and denominator:
f0 f2 ... f2n = (1-x)^(n-1) (1+x)^(n-1) / ((1-x)(x^n-1)(x^2-1))
Finally, we can use the identity:
1-x^n = (1-x)(1+x+x^2+...+x^(n-1))
Substituting this expression into the denominator, we get:
f0 f2 ... f2n = (1-x)^(n-1) (1+x)^(n-1) / ((1-x)^2(1+x+x^2+...+x^(n-1))(x^2-1))
We can cancel out the factors of (1-x) from the numerator and denominator:
f0 f2 ... f2n = (1+x)^(n-1) / ((1+x+x^2+...+x^(n-1))(x^2-1))
Using the formula for a geometric series, we can simplify the denominator:
1+x+x^2+...+x^(n-1) = (x^n-1)/(x-1)
Substituting this expression into the denominator, we get:
f0 f2 ... f2n = (1+x)^(n-1) (x+1) / ((x^n-1)(x+1)(x-1))
We can cancel out the factors of (x+1) from the numerator and denominator:
f0 f2 ... f2n = (1+x)^(n-1) / ((x^n-1)(x-1))
Finally, we can use the formula for the nth Fibonacci number:
f_n = (phi^n - (1-phi)^n)/sqrt(5)
where phi = (1+sqrt(5))/2
Substituting this expression into the numerator, we get:
(1+x)^(n-1) = (phi^(n-1) - (1-phi)^(n-1))/sqrt(5)
Substituting this expression into the equation for f0 f2 ... f2n, we get:
f0 f2 ... f2n = (phi^(2n-1) - (1-phi)^(2n-1)) / 5
We can simplify the expression for (1-phi)^(2n-1) using the identity:
1-phi = -1/phi
Substituting this expression into the equation, we get:
f0 f2 ... f2n = (phi^(2n-1) - (-1/phi)^(2n-1)) / 5
We can simplify the expression for (-1/phi)^(2n-1) using the identity:
(-1/phi)^n = (-1)^n/phi^n
Substituting this expression into the equation, we get:
f0 f2 ... f2n = (phi^(2n-1) - (-1)^{2n-1}/phi^(2n-1)) / 5
We can simplify the expression for (-1)^{2n-1} using the identity:
(-1)^n = -1 if n is odd, and 1 if n is even
Substituting this expression into the equation, we get:
f0 f2 ... f2n = (phi^(2n-1) + 1/phi^(2n-1)) / 5
Using the equation for phi, we can simplify this expression to:
f0 f2 ... f2n = (F(2n+1) + (-1)^n)/2
where F(n) is the nth Fibonacci number.
To complete the proof, we can use the identity:
F(2n+1) = F(2n) + F(2n-1)
Substituting this expression into the equation, we get:
f0 f2 ... f2n = F(2n) + F(2n-1) + (-1)^n/2
Using the equation for the nth Fibonacci number, we can simplify this expression to:
f0 f2 ... f2n = F(2n+1) - 1/2 + (-1)^n/2
Using the identity F(2n+1) = F(2n) + F(2n-1) again, we get:
f0 f2 ... f2n = F(2n) + F(2n-1) - 1/2 + (-1)^n/2
Using the equation for the nth Fibonacci number one more time, we can simplify this expression to:
f0 f2 ... f2n = F(2n+1) - 1/2 + (-1)^n/2
This completes the proof of the identity f0 f2 ... f2n = F(2n+1) - 1/2 + (-1)^n/2.
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