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#1 2012-12-18 04:43:43

jackme12
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Prove √5 is irrational...

As the subject says, can anybody prove that route 5 is irrational?

 

#2 2012-12-18 05:51:05

anonimnystefy
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Re: Prove √5 is irrational...

Let's assume that the opposite is true, that is  that square root 5 is rational. This would mean it can be written in the form p/q, where gcd(p,q)=1.

So:
sqrt(5)=p/q
Then we square both sides:
5=p^2/q^2
5*q^2=p^2
So this means that p must be divisible by 5, and we can write p=5*r and our equation becomes:
5*q^2=25*r^2
q^2=5*r^2,
But this would mean that q is also divisible by 5, and we said that p and q have a gcd of 1! This is a contradiction, so our initial assumption that sqrt(5) is rational is false!


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#3 2012-12-18 06:03:50

bob bundy
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Re: Prove √5 is irrational...

hi Stefy,

He posted again in help and I replied there.

If p^2 is divisible by 5, how do you know that p is also divisible by 5 ?

(I know it is, but I want a justification please.  Otherwise every root is irrational.)

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
 

#4 2012-12-18 06:05:17

anonimnystefy
Real Member

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Re: Prove √5 is irrational...

If p weren't divisible by 5 then neither would be p^2.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#5 2012-12-18 06:07:12

bob bundy
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Re: Prove √5 is irrational...

why?

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
 

#6 2012-12-18 06:12:28

anonimnystefy
Real Member

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Re: Prove √5 is irrational...

This can be shown by assuming that p isn't divisible by 5 and then going through 4 cases p=5k+1,5k+2,5k+3,5k+4 and squaring p in each case to see that the square of p is respectively 1, 4, 4, 1 in each case.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#7 2012-12-18 06:22:00

bob bundy
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Re: Prove √5 is irrational...

Ok.  That'll do me.  Thanks.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
 

#8 2012-12-18 06:29:19

anonimnystefy
Real Member

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Re: Prove √5 is irrational...

Don't mention it!


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#9 2012-12-18 08:42:37

jackme12
Novice

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Re: Prove √5 is irrational...

Thanks a lot, that is great help. 

Apologies about the confusion with posting in two forums.

Jack

 

#10 2012-12-18 08:50:53

anonimnystefy
Real Member

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Re: Prove √5 is irrational...

You're welome! smile


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#11 2012-12-18 22:38:41

mathaholic
Star Member

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Re: Prove √5 is irrational...

Jackme12, welcome to the forum. Sorry if I was a bit late in greeting ya. Anyways, I cannot prove yet that route 5 is irrational 'cause I am just too young to learn irrational numbers like those.
Have fun in the forum, Jackme12!

Julian


246 pages on Prime Numbers Wiki (+1)
 

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