- #1
charlie_sheep
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Hello, I’m Charlie and I’ve been trying to understand how I prove things correctly. So here is an example of the way I demonstrate:
Proof that square root of 2 isn’t rational
Consider the following hypotheses
1.(√2 is rational) → (√2 = m/n | m,n ∈ ℤ, gdc(m,n) = 1) → (m² = 2n², n = (m²)/2 | m,n ∈ ℤ, gdc(m,n) = 1)
2.(∀x,y ∈ ℤ, (x² = 2y²) → (x = 2z | z ∈ ℤ))
3.(m = 2k, n = 2p | m,n,k,p ∈ ℤ, gdc(m,n) = 1) → (gcd(m,n) = 1 ∧ gdc(m,n) ≠ 1)
4.~(gcd(m,n) = 1 ∧ gdc(m,n) ≠ 1)
By 1 and 2 there is
5.(√2 is rational) → (m = 2k, n² = 2k² | m,n,k ∈ ℤ, gdc(m,n) = 1)
By 5 and 2 there is
6.(√2 is rational) → (m = 2k, n = 2p | m,n,k,p ∈ ℤ, gdc(m,n) = 1)
By 6 and 3 there is
7.(√2 is rational) → (gcd(m,n) = 1 ∧ gdc(m,n) ≠ 1)
By 7 and 4 there is
8. ~(√2 is rational)
∎
Is it correct? If not, why? Any other observations?
Note: English is not my first language, and I’m not really good in it. So, I apologize for any mistakes.
Proof that square root of 2 isn’t rational
Consider the following hypotheses
1.(√2 is rational) → (√2 = m/n | m,n ∈ ℤ, gdc(m,n) = 1) → (m² = 2n², n = (m²)/2 | m,n ∈ ℤ, gdc(m,n) = 1)
2.(∀x,y ∈ ℤ, (x² = 2y²) → (x = 2z | z ∈ ℤ))
3.(m = 2k, n = 2p | m,n,k,p ∈ ℤ, gdc(m,n) = 1) → (gcd(m,n) = 1 ∧ gdc(m,n) ≠ 1)
4.~(gcd(m,n) = 1 ∧ gdc(m,n) ≠ 1)
By 1 and 2 there is
5.(√2 is rational) → (m = 2k, n² = 2k² | m,n,k ∈ ℤ, gdc(m,n) = 1)
By 5 and 2 there is
6.(√2 is rational) → (m = 2k, n = 2p | m,n,k,p ∈ ℤ, gdc(m,n) = 1)
By 6 and 3 there is
7.(√2 is rational) → (gcd(m,n) = 1 ∧ gdc(m,n) ≠ 1)
By 7 and 4 there is
8. ~(√2 is rational)
∎
Is it correct? If not, why? Any other observations?
Note: English is not my first language, and I’m not really good in it. So, I apologize for any mistakes.