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

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

Now, squaring the equality at the beginning we get that

[tex]3=\frac{m^2}{n^2}[/tex] (1)

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

Substituting [tex]m=3p[/tex] into (1), we get

[tex]3= \frac{9p^2}{n^2}[/tex]

which is equivalent to

[tex]n^2=3p^2[/tex].

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

Notice that

[tex]\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}[/tex].

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

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

[tex]k=\frac{k}{1}[/tex].

Then,

[tex]k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}[/tex].

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

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

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

[tex]x=q-p[/tex].

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