Let n≥ 9 be a perfect square. Then a - 1 is composite. With a equaling n.
If c (A real number) is a root of a polynomial with rational coecients,
then c is a root of a polynomial with integer coecients. (Recall that c is
said to be a root of the polynomial p(x) if p(c) = 0.)
If c (A real number) is a root of a polynomial with rational coecients,
then c is a root of a polynomial with integer coecients. (Recall that c is
said to be a root of the polynomial p(x) if p(c) = 0.)
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I stand by my answer for the first one. I realize my answer to the second was stupid.
If you have a polynomial root equation
a₀ + a₁x + a₂x² + ... + (a_n)xⁿ = 0
with rational coefficients a_i = α_i / β_i (α,β ∈ ℤ), then the solutions will be equivalent to the polynomial equation
b₀ + b₁x + b₂x² + ... + (b_n)xⁿ = 0
where b_i = a_iB, where B is the least common multiple of all the β_i. It's essentially the same thing as what you'd do to solve something like
(1/2)x + (1/5) = 0
Odds are you'd want to get rid of all the fractions, so you'd multiply through by 10, the least common multiple of the denominators to get
5x + 2 = 0
In each case you get x = -2/5, so the equations must be equivalent. You're just trying to prove the general analogue of this.
If you have a polynomial root equation
a₀ + a₁x + a₂x² + ... + (a_n)xⁿ = 0
with rational coefficients a_i = α_i / β_i (α,β ∈ ℤ), then the solutions will be equivalent to the polynomial equation
b₀ + b₁x + b₂x² + ... + (b_n)xⁿ = 0
where b_i = a_iB, where B is the least common multiple of all the β_i. It's essentially the same thing as what you'd do to solve something like
(1/2)x + (1/5) = 0
Odds are you'd want to get rid of all the fractions, so you'd multiply through by 10, the least common multiple of the denominators to get
5x + 2 = 0
In each case you get x = -2/5, so the equations must be equivalent. You're just trying to prove the general analogue of this.
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n ≥ 9 is a perfect square.
n = k² for some integer k ≥ 3
n−1 = k² − 1
n−1 = (k−1) (k+1)
Since k ≥ 3, then k−1 ≥ 2 and k+1 ≥ 4
Therefore n−1 is a composite number, since it is the product of two integer
factors that are both > 1
[I'm not too sure why we needed variable a here, since a = n]
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Let p(x) = a₀/b₀ + a₁/b₁ x + a₂/b₂ x² + ... + an/bn x^n
be a polynomial function with rational coefficient a₀/b₀, a₁/b₁, a₂/b₂, ... , an/bn
(in simplest forms) and with root c. Then p(c) = 0
Let b = least common multiple of b₀, b₁, b₂, ... , bn
Therefore b/bᵢ = some integer for all i = 0 to n, since b is a multiple of bi
Let q(x) = b p(x)
Coefficients of q(x) = b * aᵢ/bᵢ = a * (b/bᵢ) = some integer, for all i = 0 to n
So q(x) is a polynomial with integer coefficients
q(c) = b * p(c) = b * 0 = 0
c is a root of polynomial q(x), which has integer coefficients
n = k² for some integer k ≥ 3
n−1 = k² − 1
n−1 = (k−1) (k+1)
Since k ≥ 3, then k−1 ≥ 2 and k+1 ≥ 4
Therefore n−1 is a composite number, since it is the product of two integer
factors that are both > 1
[I'm not too sure why we needed variable a here, since a = n]
——————————————————————————————
Let p(x) = a₀/b₀ + a₁/b₁ x + a₂/b₂ x² + ... + an/bn x^n
be a polynomial function with rational coefficient a₀/b₀, a₁/b₁, a₂/b₂, ... , an/bn
(in simplest forms) and with root c. Then p(c) = 0
Let b = least common multiple of b₀, b₁, b₂, ... , bn
Therefore b/bᵢ = some integer for all i = 0 to n, since b is a multiple of bi
Let q(x) = b p(x)
Coefficients of q(x) = b * aᵢ/bᵢ = a * (b/bᵢ) = some integer, for all i = 0 to n
So q(x) is a polynomial with integer coefficients
q(c) = b * p(c) = b * 0 = 0
c is a root of polynomial q(x), which has integer coefficients
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for all perfect squares, a²-1 we have:
a²-1 = (a-1)(a+1). This product must be composite unless a-1=1, e.g. a=2. Thus if a²>4, it follows that a²-1 is composite.
The second problem is the contrapositive statement of Gauss's Lemma. If a polynomial is not irreducible over the rationals, then it is not irreducible over the integers.
a²-1 = (a-1)(a+1). This product must be composite unless a-1=1, e.g. a=2. Thus if a²>4, it follows that a²-1 is composite.
The second problem is the contrapositive statement of Gauss's Lemma. If a polynomial is not irreducible over the rationals, then it is not irreducible over the integers.
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Don't use Yahoo Answers and you'll get a better answer.