Question

How can u prove that sqrrt{2} is Irrational?

Answer 1

I can't explain it well. Perhaps this page will help.
http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php

Answer 2

You can't express it in the form of \(\Large \color{purple}{p \over q}\)
Can you? No, you can't.

p over q actually means that you express it in a form of a fraction. You can't in this case. You can actually, \(\Large {{sqrt{2} \over 1}}\), but this isn't really a fraction.

Answer 3

the better term is "express as quotient of two integers"

Answer 4

But I don't get it really what you are saying.Please tell me in more easy way.

Answer 5

@ParthKohli

Answer 6

I was about to type that.

Answer 7

Ya ok!

Answer 8

Assume that it is rational and then show that it cannot be expressed as the quotient of two integers.

Answer 9

If you are geometrically minded you might like this one
http://jeremykun.wordpress.com/2011/08/14/the-square-root-of-2-is-irrational-geometric-proof/

Answer 10

right ...
Let's suppose √2 were a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero. We additionally make it so that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction.


It follows that 2 = a2/b2, or a2 = 2 * b2. So the square of a is an even number since it is two times something. From this we can know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a * a would be odd too. Odd number times odd number is always odd. Check if you don't believe that!

Okay, if a itself is an even number, then a is 2 times some other whole number, or a = 2k where k is this other number. We don't need to know exactly what k is; it won't matter. Soon is coming the contradiction:

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2.


This means b2 is even, from which follows again that b itself is an even number!!!

WHY is that a contradiction? Because we started the whole process saying that a/b is simplified to the lowest terms, and now it turns out that a and b would both be even. So √2 cannot be rational.

Answer 11

But how mertsj! please tell me.

Answer 12

lol it's always fun when people suddenly post more complicated ways of solving the problem >:))) hahaha

Answer 13

@badi Lol, integers. You can put negative values too in the \(\Large {p \over q}\) thingy.

Answer 14

@lgbasallote hahahah

Answer 15

lol<:((((((((((((((((((((((((((((((((((((((((

Answer 16

Thanx! Badi

Answer 17

@maheshmeghwal9 don't complicate things in your mind.

Answer 18

this sol btw is copy-pasted from some website right :)

Answer 19

http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php COPIED FROM HERE

Answer 20

Can the asker do the problem now? I know that too many cooks spoil the broth.

Answer 21

i think i did this last year though i dont remember it now
u had to assume rt2 to be a rational no and prove that p and q were not co-prime or something

Answer 22

Ya! dude

Answer 23

many hands make light work....

Answer 24

I mean Mertsj!

Answer 25

@estudier That may be true for jobs that actually involve hands. With minds...not so much.

Answer 26

oh

Answer 27

Opposing proverbs @estudier @Mertsj

Answer 28

I don't know, community math (and other) solving seems to be growing in popularity.
Think it depends more on the nature of the problem and how good the group facilities are

Answer 29

let me share one too "if you want to live a happy life, tie it to a goal, not a person or objects ~Einstein"

Answer 30

Here is badi's solution sung by a German pupil in secondary school:
http://www.youtube.com/watch?v=vnMrUAvKtAg
Hope you enjoy it, even if you do not understand the language!