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Feliz [49]
3 years ago
12

A multiplicative inverse of 3 modulo 5 is any integer a such that 3 a ≡ 1 (mod 5) , hence for some a ∈ Z5 .

Computers and Technology
1 answer:
cricket20 [7]3 years ago
4 0
The obvious element for which it can't exist is 0 as a*0=0 independent of modulo
all other elements have an inverse:
1*1≡1
2*3≡6≡1
3*2≡6≡1
4*4≡16≡1

if there are more than a few numbers/guessing is inefficient it can be calculated using the extended euclidean algorithm
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Determine and prove whether an argument in English is valid or invalid. About Prove whether each argument is valid or invalid. F
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Each understudy on the respect roll got an A.  

No understudy who got a confinement got an A.  

No understudy who got a confinement is on the respect roll.  

No understudy who got an A missed class.  

No understudy who got a confinement got an A.  

No understudy who got a confinement missed class  

Explanation:

M(x): x missed class  

An (x): x got an A.  

D(x): x got a confinement.  

¬∃x (A(x) ∧ M(x))  

¬∃x (D(x) ∧ A(x))  

∴ ¬∃x (D(x) ∧ M(x))  

The conflict isn't considerable. Consider a class that includes a lone understudy named Frank. If M(Frank) = D(Frank) = T and A(Frank) = F, by then the hypotheses are overall evident and the end is counterfeit. Toward the day's end, Frank got a control, missed class, and didn't get an A.  

Each understudy who missed class got a confinement.  

Penelope is an understudy in the class.  

Penelope got a confinement.  

Penelope missed class.  

M(x): x missed class  

S(x): x is an understudy in the class.  

D(x): x got a confinement.  

Each understudy who missed class got a confinement.  

Penelope is an understudy in the class.  

Penelope didn't miss class.  

Penelope didn't get imprisonment.  

M(x): x missed class  

S(x): x is an understudy in the class.  

D(x): x got imprisonment.  

Each understudy who missed class or got imprisonment didn't get an A.  

Penelope is an understudy in the class.  

Penelope got an A.  

Penelope didn't get repression.  

M(x): x missed class  

S(x): x is an understudy in the class.  

D(x): x got a repression.  

An (ax): x got an A.  

H(x): x is on the regard roll  

An (x): x got an A.  

D(x): x got a repression.  

∀x (H(x) → A(x)) a  

¬∃x (D(x) ∧ A(x))  

∴ ¬∃x (D(x) ∧ H(x))  

Real.  

1. ∀x (H(x) → A(x)) Hypothesis  

2. c is a self-self-assured element Element definition  

3. H(c) → A(c) Universal dispatch, 1, 2  

4. ¬∃x (D(x) ∧ A(x)) Hypothesis  

5. ∀x ¬(D(x) ∧ A(x)) De Morgan's law, 4  

6. ¬(D(c) ∧ A(c)) Universal dispatch, 2, 5  

7. ¬D(c) ∨ ¬A(c) De Morgan's law, 6  

8. ¬A(c) ∨ ¬D(c) Commutative law, 7  

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13. ∀x ¬(D(x) ∧ H(x)) Universal hypothesis, 2, 12  

14. ¬∃x (D(x) ∧ H(x)) De Morgan's law, 13  

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