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Helen [10]
2 years ago
9

Prove that 41 is congruent to 21 (mod 3). Explain using words, symbols, as you wish

Mathematics
1 answer:
Verdich [7]2 years ago
6 0

From the proof of modular congruence below, it has been shown that;

41 ≡ 21 (mod 3).

<h3>How to Solve Modular Arithmetic?</h3>

We want to use the definition of modular congruence to prove that;

41 is congruent to 21 (mod 3) i.e if a ≡ b (mod m) then b ≡ a (mod m).

We are trying to prove that modular congruence mod 3 is a symmetric relation on the integers.

First, if we recall the definition of modular congruence:

For integers a, b and positive integer m,  

a ≡ b (mod m) if and only if m|a–b

Suppose 41 ≡ 21 (mod 3).

Then, by definition, 3|41–21, so there is an integer k such that 41 – 21 = 3k.

Thus;

–(41 – 21) = –3k

So

21 – 41 = 3(–k)

This shows that 3|21 – 41.

Thus;

21 ≡ 41 (mod 3) and the proof is complete

Read more about Modular Arithmetic at; brainly.com/question/16032865

#SPJ1

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.
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Answer:

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Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then  

\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

\bf G(x,y) = (x, y, 3-x^2-y^2)

with 0≤ x≤1 and  0≤ y≤1

So

\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)

we also have

\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)

and so

\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and  0≤ y≤1

\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

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3 years ago
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