Question

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

Answer 1

1. Let us proof that √3 is an irrational number, using reductio ad absurdum. Assume that
`√(3)=(m)/(n)` where
`m` and
`n` are non negative integers, and the fraction
`(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=(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= (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

`(m)/(n) = (3p)/(3q) = (p)/(q)`.

The above equality means that the fraction
`(m)/(n)` is reducible, what contradicts our initial assumption. So,
`√(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=(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=(k)/(1)`.

Then,

`k\cdot r = (k)/(1)(m)/(n) = (mk)/(n)`.

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

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
`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.