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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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