Direct Proof: Every odd integer is a difference of two squares

 

 

 

 

 

 

1. Direct Proof: Every odd integer is a difference of two squares.
Let’s consider an odd integer n.
An odd integer can be represented as 2k + 1, where k is an integer.
Now, we aim to express n as the difference of two squares: m^2 – (m-1)^2.

n = m^2 – (m-1)^2
n = m^2 – (m^2 – 2m + 1)
n = 2m – 1

Let k = m – 1, then n = 2k + 1, which is the definition of an odd integer.
Therefore, every odd integer is indeed a difference of two squares.

2. Direct Proof: If a^2 is not divisible by 4, then a is odd.
Let’s prove by contrapositive:
Assume a is even, then a = 2k for some integer k.
a^2 = (2k)^2 = 4k^2, which is divisible by 4.

Therefore, if a^2 is not divisible by 4, then a must be odd.

3. False Statement: The number log2(3) is rational.
Proof by contradiction:
Assume log2(3) is rational, then it can be expressed as a ratio of two integers: log2(3) = p/q where p and q are integers.
This implies 2^(p/q) = 3, leading to 2^p = 3^q, which is not possible since 2 and 3 are prime numbers.

Hence, log2(3) is irrational.

4. Direct Proof: If ab is odd, then a^3 + b^3 is even.
Let’s consider a and b to be odd integers.
a^3 + b^3 = (a + b)(a^2 – ab + b^2)
Since a and b are odd, a + b is even.
The product of an even number and any number (a^2 – ab + b^2) results in an even number.

Thus, if ab is odd, then a^3 + b^3 is indeed even.

5. Set Theory Proof: A ∪ B ∩ B ∪ A = A
Let x be an element of A ∪ B ∩ B ∪ A.
This implies x ∈ A or x ∈ B and x ∈ B or x ∈ A.
Therefore, x ∈ A or x ∈ B, which means x ∈ A.
Hence, A ∪ B ∩ B ∪ A = A.

6. Direct Proof: If X ⊆ A ∩ B, then X ⊆ A or X ⊆ B.
Let x be an element of X.
Since X ⊆ A ∩ B, x ∈ A and x ∈ B.
Therefore, x ∈ A or x ∈ B, which implies X ⊆ A or X ⊆ B.

7. Direct Proof: Prove that 9 | (4^(3n) + 8) for every integer n ≥ 0.
Let’s express the expression as 4^(3n) + 8 = (4^3)^n + 8 = 64^n + 8.
64^n can be written as (63 + 1)^n = Sum(k=0 to n)((n choose k)*63^(n-k)).
This expansion includes all terms divisible by 9 except the last term, which is 1.
Thus, 64^n + 8 is divisible by 9 for every integer n ≥ 0.

8. Proof using Mathematical Induction: Σ k=1 to n k^2 = ((n(n+1))/2)^2
Base Case: For n = 1, Σ k=1 to 1 k^2 = (1(1+1)/2)^2 = (1)^2 = 1.
Inductive Step: Assume true for n = m: Σ k=1 to m k^2 = (m(m+1)/2)^2.
Show for n = m+1:
Σ k=1 to m+1 k^2 = Σ k=1 to m k^2 + (m+1)^2
= (m(m+1)/2)^2 + (m+1)^2
= ((m+1)(m+2)/2)^2

Therefore, by induction, Σ k=1 to n k^2 = ((n(n+1))/2)^2 for all positive integers n.

9. Proof using Product Notation: ∏ n N (12n + 4)/(12n + 2)
∏ n N (12n + 4)/(12n + 2) = (4/2)(16/14)(28/26)…((12n+4)/(12n+2))
= ((12n+4)/(12n+2))*(12(n+1)+4)/(12(n+1)+2)
= (12(n+1)+4)/(12(n+1)+2)

Therefore, for all positive integers n, the product formula holds true.