Question

proof that square root of 2 is irrational in a simple explanation PLEASE

Answer 1

there are many. google and choose

Answer 2

please explain it to me. in a simple way. can't really understand how to prove this

Answer 3

all proofs start out the same. assume it is rational and arrive at a contradition

Answer 4

in other words assume
\[\sqrt{2}=\frac{a}{b}\] where a and b are integers. that is the first step of every proof

Answer 5

the next step is to say that this means that
\[2=\frac{a^2}{b^2}\] and so
\[2b^2=a^2\]

Answer 6

then?

Answer 7

that is the next step of every proof. you should understand these steps before you continue

Answer 8

now you have several choices. you can start arguing about even and odd until you mind goes numb

Answer 9

alfers101 i don't mean to intrude, but i think you should pay more attention in your math class... ALL your hw gets solved here and you don't seem to be getting any of it... it's better to understand things so you can do them by yourself

Answer 10

think about problems such a statement can cause....if say you were to encounter it as part of problem.
Hint: checking domains is usually a good place to start

Answer 11

Irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers, with b non-zero, and is therefore not a rational number. Informally, this means that an irrational number cannot be represented as a simple fraction.

The square roots of all numbers which are not perfect squares are irrational and a proof may be found in quadratic irrationals.
The irrationality of the square root of 2 may be proved by assuming it is rational and inferring a contradiction, called an argument by reductio ad absurdum. The following argument appeals twice to the fact that the square of an odd integer is always odd.


If √2 is rational it has the form m/n for integers m, n not both even. Then m2 = 2n2, hence m is even, say m = 2p. Thus 4p2 = 2n2 so 2p2 = n2, hence n is also even, a contradiction.

Answer 12

okay i'll try my very best! Good luck to me..

Answer 13

infauzan has the even odd proof. if you like that, fine. so long as you understand it

Answer 14

or you can just look at the statement
\[2b^2=a^2\] and think that if you factor a into the product of primes, and then square, all the prime powers will be even (raise to the power of two means double) and likewise if you factor b into the product of primes and square all the exponents will also be even. and therefore the power of 2 on the right is even but the power of two on the left is odd because of the 2 out front.

Answer 15

this is really the same as the usual one but it extends to other numbers than 2. for example you could argue in the same way that
\[\sqrt{3}\] is irrational by writing
\[3b^2=a^2\] and making the exact same arguement

Answer 16

well, this is the truth. i really love math! i understand the trigonometry,geometry,staticstics very well. but algebra is the only one that i can't understand! i don't know why. but i'm still i'm doing my best to solve there.

Answer 17

thanks for the help