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Lorico [155]
3 years ago
8

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

is a multiple of thek is a multiple of 3. Hint: The proof is very similar to the proof that is inational 5. Use a direct proof to show that the product of a rational number and an integer must be a rational number 6 Use a proof by contradiction to show that the sum of an integer and animational number must be irrational
Mathematics
1 answer:
Ierofanga [76]3 years ago
7 0

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that \sqrt{3}=\frac{m}{n} where  m and n are non negative integers, and the fraction \frac{m}{n} is irreducible, i.e., the numbers m and n have no common factors.

Now, squaring the equality at the beginning we get that

3=\frac{m^2}{n^2} (1)

which is equivalent to 3n^2=m^2. From this we can deduce that 3 divides the number m^2, and necessarily 3 must divide m. Thus, m=3p, where p is a non negative integer.

Substituting m=3p into (1), we get

3= \frac{9p^2}{n^2}

which is equivalent to

n^2=3p^2.

Thus, 3 divides n^2 and necessarily 3 must divide n. Hence, n=3q where q is a non negative integer.

Notice that

\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}.

The above equality means that the fraction \frac{m}{n} is reducible, what contradicts our initial assumption. So, \sqrt{3} is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, r\in\mathbb{Q}, which is equivalent to say that r=\frac{m}{n} where  m and n are non negative integers. Also, assume that k\in\mathbb{Z}. So, we want to prove that k\cdot r\in\mathbb{Z}. Recall that an integer k can be written as

k=\frac{k}{1}.

Then,

k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}.

Notice that the product mk is an integer. Thus, the fraction \frac{mk}{n} is a rational number. Therefore, k\cdot r\in\mathbb{Q}.

3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have x is irrational and p\in\mathbb{Q}.

Write q=x+p and let us suppose that q is a rational number. So, we get that

x=q-p.

But the subtraction or addition of two rational numbers is rational too. Then, the number x must be rational too, which is a clear contradiction with our hypothesis. Therefore, x+p is irrational.

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Find the area of a regular hexagon if given a side length of 1.9 to the square root of 3.
dalvyx [7]

Answer:

28.137 sq. units.

Step-by-step explanation:

Area of a regular hexagon is

A=\dfrac{3\sqrt{3}}{2}a^2

where, a is the side length.

It is given that the side length of the regular hexagon is 1.9\sqrt{3}.

Substitute a=1.9\sqrt{3} in the above formula.

A=\dfrac{3\sqrt{3}}{2}(1.9\sqrt{3})^2

A=(1.5\sqrt{3})(10.83)

A=16.245\sqrt{3}

A=28.137

Therefore, the area of regular hexagon is 28.137 sq. units.

6 0
3 years ago
Use the elimination method to solve the system of equations
Lady bird [3.3K]

Answer:

D

Step-by-step explanation:

2x + 3y = 18

3x - 3y = 12

the 3y and -3y cancel and you add the other like terms

5x = 30

x = 6

plug x back into any of the equations

2(6) + 3y = 18

12 + 3y = 18

3y = 6

y = 2

answer is (6,2)

4 0
2 years ago
I need help please?!!!!!
ozzi

Answer:

8

Step-by-step explanation:

plug in the number 4, it would be 5•4-12

Do 5•4=20

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2 years ago
Divide ₹ 60 in the ratio 1 : 2 between Kriti and Kiran.​
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Step-by-step explanation:

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6 0
3 years ago
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If z varies inversely as w, and z= 3 when w=4, find z when w=2.<br> Z=
vivado [14]

Answer:

6

Step-by-step explanation:

Inverse proportion.

z = k/w

z = 3, w = 4

Solve for k constant.

3 = k/4

(4)3 = (4)k/4

12=k

The value of k constant is 12.

z = k/w

Put w as 2 and k as 12 and solve for z.

z = 12/2

z = 6

The value of z when w equals 2 is 6.

8 0
3 years ago
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