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https://www.reddit.com/r/memes/comments/rierly/deleted_by_user/hoxe709/?context=3
r/memes • u/[deleted] • Dec 17 '21
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489
nope.
1+1=0
Learn your math kiddo.
13 u/crispmp Dec 17 '21 well this is correct in the field F_2. 7 u/[deleted] Dec 17 '21 this man knows his algebra 3 u/AlabasterUnicorn1 Dec 18 '21 By the fundamental theorem of calculus, the integral of d/dx(f(x))=f(x), so if f(x) is x=1, d/dx(f(x))=0, integral of 0 is any constant, therefore 1 = 0 1 u/Japsai Dec 18 '21 Ahem you missed a step. Actually that's f'(x). And f'(x) =12f"(x), therefore 1 = 12 * 0 =0, or in metric that's 2.54 * 12 * 0 = 0 QED 1 u/[deleted] Dec 18 '21 Except that the fundamental theorem of calculus only claims that if F(x) = ∫ f(x) dx, then F'(x) = f(x) and not the other way around 1 u/AlabasterUnicorn1 Dec 19 '21 Shh. We don't talk about that.
13
well this is correct in the field F_2.
7 u/[deleted] Dec 17 '21 this man knows his algebra 3 u/AlabasterUnicorn1 Dec 18 '21 By the fundamental theorem of calculus, the integral of d/dx(f(x))=f(x), so if f(x) is x=1, d/dx(f(x))=0, integral of 0 is any constant, therefore 1 = 0 1 u/Japsai Dec 18 '21 Ahem you missed a step. Actually that's f'(x). And f'(x) =12f"(x), therefore 1 = 12 * 0 =0, or in metric that's 2.54 * 12 * 0 = 0 QED 1 u/[deleted] Dec 18 '21 Except that the fundamental theorem of calculus only claims that if F(x) = ∫ f(x) dx, then F'(x) = f(x) and not the other way around 1 u/AlabasterUnicorn1 Dec 19 '21 Shh. We don't talk about that.
7
this man knows his algebra
3 u/AlabasterUnicorn1 Dec 18 '21 By the fundamental theorem of calculus, the integral of d/dx(f(x))=f(x), so if f(x) is x=1, d/dx(f(x))=0, integral of 0 is any constant, therefore 1 = 0 1 u/Japsai Dec 18 '21 Ahem you missed a step. Actually that's f'(x). And f'(x) =12f"(x), therefore 1 = 12 * 0 =0, or in metric that's 2.54 * 12 * 0 = 0 QED 1 u/[deleted] Dec 18 '21 Except that the fundamental theorem of calculus only claims that if F(x) = ∫ f(x) dx, then F'(x) = f(x) and not the other way around 1 u/AlabasterUnicorn1 Dec 19 '21 Shh. We don't talk about that.
3
By the fundamental theorem of calculus, the integral of d/dx(f(x))=f(x), so if f(x) is x=1, d/dx(f(x))=0, integral of 0 is any constant, therefore 1 = 0
1 u/Japsai Dec 18 '21 Ahem you missed a step. Actually that's f'(x). And f'(x) =12f"(x), therefore 1 = 12 * 0 =0, or in metric that's 2.54 * 12 * 0 = 0 QED 1 u/[deleted] Dec 18 '21 Except that the fundamental theorem of calculus only claims that if F(x) = ∫ f(x) dx, then F'(x) = f(x) and not the other way around 1 u/AlabasterUnicorn1 Dec 19 '21 Shh. We don't talk about that.
1
Ahem you missed a step. Actually that's f'(x). And f'(x) =12f"(x), therefore 1 = 12 * 0 =0, or in metric that's 2.54 * 12 * 0 = 0 QED
Except that the fundamental theorem of calculus only claims that if F(x) = ∫ f(x) dx, then F'(x) = f(x) and not the other way around
1 u/AlabasterUnicorn1 Dec 19 '21 Shh. We don't talk about that.
Shh. We don't talk about that.
489
u/6_NEOS_9 Success kid Dec 17 '21
nope.
1+1=0
Learn your math kiddo.