Question

prove that 3+√2 is not a irrational number , given that √2 is not rational​

Answer 1

Answer:

Step-by-step explanation:

From the equality √2  = a/b 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 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 it if you don't believe me!

Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes 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

Answer 2

Answer:

the question is wrong because √2 is irrational number the question is :

prove that 3+√2 is an irrational number , given that √2 is an irrational number

Step-by-step explanation:

OR

if this is only your question then the answer is :

let us assume that 3+√2 is irrational number

3+√2 = p/q { where p and q may equal to 0}

√2 = -3

here -3 is rational number and it is given that √2 is not irrational so both LHS and RHS are not irrational so the number is 3+√2 is also not irrational

Only this is the way we can prove it