Rational or not?

3 + 12 27 \sqrt{3}+\sqrt{12}-\sqrt{27} Is the number above rational or irrational?

Rational Irrational

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

6 solutions

Áron Bán-Szabó
Jul 20, 2017

3 , 12 , 27 3, 12, 27 Note that all of the numbers above are multiple of 3 3 .

12 = 3 4 = 3 4 = 2 3 27 = 3 9 = 3 9 = 3 3 \begin{aligned} \sqrt{12} & =\sqrt{3*4} \\ & =\sqrt{3}*\sqrt{4} \\ & = 2\sqrt{3} \\ \\ \sqrt{27} & = \sqrt{3*9} \\ & = \sqrt{3}*\sqrt{9} \\ & =3\sqrt{3} \end{aligned}

So 3 + 12 27 = 3 + 2 3 3 3 = 3 3 3 3 = 0 \sqrt{3}+\sqrt{12}-\sqrt{27}=\sqrt{3}+2\sqrt{3}-3\sqrt{3}=3\sqrt{3}-3\sqrt{3}=\boxed{0}

Therefore the answer is rational.

Nice problem and solution! (+1)

Noel Lo - 3 years, 10 months ago
Mohammad Khaza
Sep 11, 2017

3 + \sqrt3+ 12 \sqrt{12} 2 7 -\sqrt27

= 3 + \sqrt3+ 2 3 2\sqrt3 3 3 -3\sqrt3

= 3 3 3\sqrt3- 3 3 3\sqrt3

= 0 0

Arthur Conmy
Jul 21, 2017

Relevant wiki: Surds

3 + 12 27 \sqrt{3}+\sqrt{12}-\sqrt{27} .

= 3 ( 1 + 4 9 ) =\sqrt{3}(1+\sqrt{4}-\sqrt{9}) .

= 3 ( 1 + 2 3 ) =\sqrt{3}(1+2-3) .

= 3 ( 0 ) =\sqrt{3}(0) .

= 0 =0 .

... and 0 is r a t i o n a l \boxed{rational} :D

Christian Daang
Jul 20, 2017

The answer is 0, which is a rational number. :D


I think, my reasoning is so bad ( I can't explain it completely (like for instance, using some lemma, and stuffs) but, uhm, what if the question is

"Is i i i - i an imaginary (complex) number or real number ? "

What is the answer then? :)

Are you saying you don't know how to simplify "i - i" ?

Pi Han Goh - 3 years, 10 months ago

Log in to reply

Nope sir, it's just you know, like illogical to subtract an imaginary number to another imaginary number. XD (they are both imaginary, I am just thinking that, is the operation of subtraction really applies to imaginary (complex) numbers ? Is there a lemma about it? )

Christian Daang - 3 years, 10 months ago

N Z Q R C \mathbb{N}\subseteq\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}\subseteq\mathbb{C}

i i = 0 i-i=0 is real, therefore it is also complex (and a member of all the above sets) because they are subsets of each other.

This problem talks about rationals and irrationals, which are complements (I believe). All rationals (by definition) are rational, and by the subset property therefore all integers and naturals are as well.

Hope you can understand!

Arthur Conmy - 3 years, 10 months ago

Log in to reply

I see it now sir. XD Thanks. ^^

Christian Daang - 3 years, 10 months ago

Log in to reply

I am 16, like you. No need for 'sir' :)

Arthur Conmy - 3 years, 10 months ago

Log in to reply

@Arthur Conmy Oh, I didn't notice that, sorry. :D

Christian Daang - 3 years, 10 months ago
Chandresh Shah
Nov 8, 2017

root 3 + 2root 3 - 3root 3 =3root3-3root 3= 0, Zero is rational number.

Kazi Sifat
Jul 30, 2017

Ama puimama pumapimu mami wama puma wama.uei23746578283

Rutry466637264577573

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...