A better proof for the set of irrational number not closed under ordinary multiplication.

Question

A positive irrational number $$q$$ is by definition a real number than cannot be expressed as a ratio of $2$ integers. To show that the set of irrational number is not closed under ordinary multiplication, I seek a counter-example that is $$\sqrt{2} \times \sqrt{2} = 2 = \frac{2}{1}$$ which is obvious as can be seen that the product of $2$ irrational number is a positive rational number which is not in the set of positive irrational number.

Here is my two Questions

1) How do I notate the set of irrational numbers?

2) How do I show the above proof symbolically?

Answer 1

Usually, the set of irrational numbers is written simply as $\mathbb R\setminus \mathbb Q$.

As for the symbolic proof, my advice is "Avoid symbolic proofs."

A good proof in mathematics is not one that is fully written with symbols alone. A good proof is written in words, but is still mathematically rigorous, such that there is no doubt that any mathematitian, if he really set his mind to it, could write it down symbolically.

The proof, written in mathematical language, would go like this:

We wish to prove that $S=\mathbb R\setminus \mathbb Q$ is not closed for multiplication. A set is closed for multiplication if:

$$\forall x,y\in S: x\cdot y\in S$$

This means the set is not closed for multiplication if $$\exists x,y\in S: x\cdot y\notin S$$

In our case, let $x=y=\sqrt 2$. Obviously, $x,y\in S$, since $\sqrt 2 \notin \mathbb Q$. Then, we can see that

$$x\cdot y = \sqrt{2} \cdot \sqrt 2 = 2 \in \mathbb Q,$$ meaning that $x\cdot y\notin S$. This proves that $S$ is not closed for multiplication.

Answer 2

You denote (not notate!) the set of rational numbers by $\mathbb Q$ and that of the real numbers by $\mathbb R$ ; therefore the set of irrational numbers can be written as $\mathbb R \backslash \mathbb Q$ or $\mathbb R - \mathbb Q$, depending on your taste. Your proof could simply go as follows : since $\sqrt 2 \in \mathbb R \backslash \mathbb Q$ but $(\sqrt 2)^2 = 2 \in \mathbb Q$, $\mathbb R \backslash \mathbb Q$ is not closed under multiplication.

Hope that helps,

Answer 3

Now the challenge is that not all numbers under a root is irrational. For example, $$\sqrt{4} = 2, \sqrt{9} = 3, \sqrt{16} = 4, ..... $$ are all rational numbers

However, all the numbers under the square root between these numbers are irrational. For example, $$\sqrt{5}, \sqrt{6}, \sqrt{7},\sqrt{8}$$ are all irrational numbers and of course the most popular irrational number $\large{\color{red}{\pi}}$.

If you notice you will see that all the rational numbers under the root in the first list are all square numbers $$4=2^2,9=3^2,16=4^2$$ and so here is how you can prove it symbolically, Let $q$ be a non square positive number, and so $\sqrt{q}$ is irrational, now observe also that $\frac{1}{\sqrt{q}}$ is also irrational, and now if we multiply those together $$ \sqrt{q} \times \frac{1}{\sqrt{q}} = 1$$ which is obviously rational and so , The multiplication of two irrational numbers is not necessarily an irrational number as we saw.

There is no standard notation for the set of irrational numbers, but the notations $\bar{\mathbb{Q}}$, $\mathbb{R-Q}$, or $\mathbb{R \backslash Q}$, where the $\bar{}$, minus sign, or backslash indicates the set complement of the rational numbers Q over the reals R, could all be used.

Answer 4

Here is an alternative argument which is almost as simple as yours but more general; e.g., it also shows that the set of transcendental numbers is not closed under multiplication.

Theorem. If $X$ is a subset of $\mathbb R$ such that
(a) the complement $\mathbb R\setminus X$ is countable, and
(b) there is a nonzero number $c\in\mathbb R\setminus X$,
then $X$ is not closed under multiplication.

Proof. Consider the hyperbola $S=\{(x,y)\in \mathbb R\times\mathbb R:xy=c\}.$ Since the projections of $S$ onto the horizontal and vertical axes are one-to-one maps, $\{(x,y)\in S:x\in\mathbb R\setminus X\text{ or }y\in\mathbb R\setminus X\}$ is countable. Since $S$ is uncountable, there is a point $(x,y)\in S$ such that $x\in X$ and $y\in X,$ while $xy=c\notin X.$

(Of course the assumption that $\mathbb R\setminus X$ is countable could be weakened to "$\mathbb R\setminus X$ has cardinality less than the continuum.")

Answer 5

If a positive integer is not a perfect squae then its square root is irrational: So take a perfect square, say $100$ and factorize it into two numbers which are not perfect squares, say $100=5\times20$. Now take square roots on both sides of the equation. We get $10=\sqrt{100}=\sqrt{5}\times\sqrt{20}$, the last two are irrationals with an integer as their product.

Answer 6

i got a good one, let say for multiplication.

let say that r is retional and i=√2 is irretional. now let say that r*i is retional. that means that, for some integars m and n. m/n will be the most reduced fraction possible.

m/n=r*i if we square everything we get that

m^2/n^2=r^2*2 from that we can see that m have to be an even number, lets call it 2k=m. means that: 4k^2=r^2*n^2*2<->2k^2=r^2*n^2 which means that r^2*n^2 have to be even, therefor n must be even and that can be because we choosed a reduced fraction!