Question

Prove √2 as irrational.
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Answer 1

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/b22=4k2/b22*b2=4k2b2=2k2.


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

Answer 2

Answer:

Step-by-step explanation:

let us assume that √2  as irrational

so ,

√2 = a/b

b√2 = a

squaring on both sides  , we get

2b² = a² -------------------------------------------------------- ( 1  )

if 2 divides a² then it also divides a

so ,

a = 2c

squaring on both sides , we get

a² = 4c²---------------------------------------------------------(  2  )

substitute equation( 1 )in equation (2 )

we get

2b² = 4 c²

b² = 2 c²

here a b and c have 2 as common factor

so 2 is rational number

but ,

√2 is irrational

this contradiction has arisen due to our incorrect assumption

this contradicts the fact that ,

√2 is a irrational number

hope it helps u