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$$\frac{\sqrt{2}}{2}$$ 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?

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    $\begingroup$ Hint: Product of rational numbers is again rational. $\endgroup$ – Krish Mar 23 '15 at 11:04
  • $\begingroup$ You have changed the clue. OK then. $\endgroup$ – Sufyan Naeem Mar 23 '15 at 11:08
  • $\begingroup$ It's definitely irrational, but it's not transcendental. It's an algebraic number but it's not an algebraic integer (minimal polynomial is $2x^2 - 1$). $\endgroup$ – user153918 Mar 23 '15 at 13:43
  • $\begingroup$ See this answer for the general ideal behind this (complement of subgroup closure). $\endgroup$ – Bill Dubuque Mar 23 '15 at 14:39
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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.

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  • $\begingroup$ How you taken fraction of $p$ and $q$ equal to $\frac{\sqrt2}{2}$ however the quotient of two fractions you stated can't be equal. One is rational and one is irrational i.e. $\frac{\sqrt2}{2}$? $\endgroup$ – Sufyan Naeem Mar 23 '15 at 11:58
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    $\begingroup$ I used a method called "proof by contradiction". You suppose something is true, use logic to arrive at a contradiction, and conclude that the assumed thing must in fact be false. In this case, I assumed that $\sqrt(2)/2$ was rational...which led to a contradiction. So the assumption is false, and we see that $\sqrt{2}/2$ must actually be rational. $\endgroup$ – John Hughes Mar 23 '15 at 12:22
  • $\begingroup$ Alright then, I must accept your answer now however this is not an unique answer but you helped me out at my confusions with your comment. :) $\endgroup$ – Sufyan Naeem Mar 23 '15 at 12:27
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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).

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  • $\begingroup$ Since you said that product of rational and irrational is irrational then you let $q=px$ and then dividing by $p$, you got the equation in the form $\frac{q}{p}$. I don't think it clears my query. Since I can't understand how you assumed $q$ as an integer however it's an irrational number. How can an irrational number be an integer? $\endgroup$ – Sufyan Naeem Mar 23 '15 at 11:53
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    $\begingroup$ @SufyanSheijk I assumed that $px$ was rational, and I give the name $q$ to whichever rational number that happens to be. This leads to a contradiction, meaning that the assumption must have been wrong. $\endgroup$ – Arthur Mar 23 '15 at 18:15
  • $\begingroup$ Yeah I know it.. Alright. $\endgroup$ – Sufyan Naeem Mar 23 '15 at 20:03
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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!

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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.

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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

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    $\begingroup$ very kind of you! yes, i was just trying to be helpful wrt notation. the answer to your specific question follows directly from the definition: $x$ is irrational if it cannot be expressed as the ratio of two integers. if $\frac{p}{q}x$ (for integers $p$ and $q$ were rational then $\exists m,n \in \mathbb{Z}$ with $\frac{p}{q}x=\frac{m}{n}$. this gives $x= \frac{mq}{np}$ which contradicts the assumed irrationality of $x$ - (in fact Emilio's answer points this out clearly) $\endgroup$ – David Holden Mar 23 '15 at 11:26
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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.

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