
#1
Sep805, 09:11 PM

P: 22

This is Algebra 2 question...
I have to prove that the square root of 2 is irrational... First we must assume that sqrt (2) = a/b I never took geometry and i dont know proofs... Please help me. Thank you. 



#2
Sep805, 09:33 PM

P: 159

a rational number is of form a/b where a and be are mutually prime. I will give you a hint: you must prove that and and b cannot possibly be mutually prime.
and what does this have to do with geometry? 



#3
Sep805, 10:12 PM

P: 669

You're off to a good start. Let "a" and "b" be natural numbers.
[tex]\sqrt{2}=\frac{a}{b}\implies b\sqrt{2}=a[/tex] Now, how could "a" be a natural number? Well, "b" would be some multiple of [itex]\sqrt{2}[/itex]. This in turn, would mean that "b" isn't a natural number. Can you see where this is going? You need to prove this by contradiction. 



#4
Sep905, 06:02 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

Prove the Square Root of 2 is irrational
"Now, how could "a" be a natural number? Well, "b" would be some multiple of [itex]\sqrt{2}[/itex]. This in turn, would mean that "b" isn't a natural number."
How does that follow? Saying b*(1/2) , for example, equals a natural number does not imply that b isn't a natural number! Of course 1/2 isn't an irrational number but the whole point here is to prove that [itex]\sqrt{2}[/itex] is irrational. Better to note that if [itex]\frac{a}{b}= \sqrt{2}[/itex] then, squaring both sides, [itex]\frac{a^2}{b^2}= 2[/itex] so that a^{2}= 2b^{2} showing that a^{2} is even. Crucial point: the square of an odd integer is always odd: If p is an odd integer, then it can be written 2n+ 1 where n is any integer. p^{2}= (2n+1)^{2}= 4n^{2}+ 4n+ 1= 2(2n^{2}+2n)+1. Since 2n^{2}+ 2n is an integer, p^{2} is of the form 2m+1 (m= 2n^{2}+2n) and so is odd. Do you see why knowing that tells us that a must be even? 



#5
Sep905, 01:50 PM

P: 22




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