What type of number is this $\frac{\sqrt2}{2}$?

Question

$$\frac{\sqrt{2}}{2}$$ In this monomial, an irrational number is divided by a rational number. However this is not a general case but can any one tell me that when we divide an irrational number or multiply an irrational number or its multiplicative inverse by a rational number then what type of the number we get in output? Either rational or irrational?

Answer 1

Suppose that $$\frac{\sqrt{2}}{2}= \frac{p}{q}$$ with $p$ and $q$ integers. Multiply through by $2$ to get $$\sqrt{2}= \frac{2p}{q}$$.

That would make $\sqrt{2}$ be rational, which it's not. So the assumption that $\frac{\sqrt{2}}{2}$ is rational must be wrong.

More generally: irrational divided by rational is still irrational.

Answer 2

Let $a$ irrational and suppose $b=\dfrac{n}{m}a$ ( $n,m \in \mathbb{Z}$) be rational i.e. $b= \dfrac{p}{q}$ ( $p,q \in \mathbb{Z}$).

You have: $$ \dfrac{p}{q}=\dfrac{n}{m}a \Rightarrow a=\dfrac{pm}{qn} $$ that is rational : contradiction!

Answer 3

Suppose that $\frac{\sqrt{2}}{2}=\frac{p}{q}$, then $\sqrt{2}=\frac{2p}{q}$, hence $\sqrt{2}$ would be rational, this is clearly a contradiction.

Answer 4

If $p$ is a rational (non-zero) number, and $x$ is irrational, then $$ px, \quad \frac px, \quad \frac xp, \quad p+x, \quad p-x $$ are all irrational. This can be proven by contradiction. Assume, for instance, that $px$ is a rational number, call it $q$. That is, set $q = px$. Now divide through by $p$, and we get $$\frac{q}{p} = x$$ The left-hand side is a fraction of two rational numbers, so it is rational. The right-hand side is $x$, so it is irrational. These two can therefore not be equal, and therefore the assumption that $px$ is rational must be false.

The others can be proven the same way. Be careful, however, because for each of the five expressions $$ xy, \quad \frac xy, \quad \frac yx, \quad x+y, \quad y-x $$there are some irrational numbers $x, y$ such that makes it rational (although not all at once).

Answer 5

be careful with terminology! the same number is the denotation of many different expressions, e.g. $$ \frac{\sqrt{2}}{2} = \frac1{\sqrt{2}} = 2^{-\frac12} = \sin \frac{\pi}4 $$ none of these is properly described as a polynomial btw

Answer 6

If you make the product of an irrational number $a$ with a non-null rational number $b$ then you will always get an irrational number, because :

$$a=(ab)\times b^{-1} $$

So if $ab$ happens to be a rational number then, because $b^{-1}$ is a rational number you end up by saying $a$ is a rational number (because a product of rational numbers is always a rational number), which is false.

However there is no general rule if you ask irrational $\times$ irrational because :

$$\sqrt{2}\times \frac{1}{\sqrt{2}}=1\text{ is rational} $$

$$\sqrt{2}\times \sqrt{3}=\sqrt{6}\text{ is irrational}$$

The main point is that $\mathbb{Q}$ is stable under multiplication and inverse of non-null elements (i.e. it is a field).

You have exactly the same thing appearing when considering algebraic numbers vs. transcendant numbers.