Answer:
what are the followings..........
Yes, 23 has an inverse mod 1000 because gcd(23, 1000) = 1 (i.e. they are coprime).
Let <em>x</em> be the inverse. Then <em>x</em> is such that
23<em>x</em> ≡ 1 (mod 1000)
Use the Euclidean algorithm to solve for <em>x</em> :
1000 = 43×23 + 11
23 = 2×11 + 1
→ 1 ≡ 23 - 2×11 (mod 1000)
→ 1 ≡ 23 - 2×(1000 - 43×23) (mod 1000)
→ 1 ≡ 23 - 2×1000 + 86×23 (mod 1000)
→ 1 ≡ 87×23 - 2×1000 ≡ 87×23 (mod 1000)
→ 23⁻¹ ≡ 87 (mod 1000)
Answer:
See Explanation
Step-by-step explanation:
(a) Proof: Product of two rational numbers
Using direct proofs.
Let the two rational numbers be A and B.
Such that:


The product:




Proved, because 1/3 is rational
(b) Proof: Quotient of a rational number and a non-zero rational number
Using direct proofs.
Let the two rational numbers be A and B.
Such that:


The quotient:

Express as product



Proved, because 3/4 is rational
(c) x + y is rational (missing from the question)
Using direct proofs.
Let x and y be
Such that:


The sum:

Take LCM


Proved, because 7/6 is rational
<em>The above proof works for all values of A, B, x and y; as long as they are rational values</em>
I need help my self lol XD sorryyyyyyyyy88898