Question

The square root of 2 is a rational number.
True
False

Answer 1

false

Answer 2

One proof of the number's irrationality is the following proof by infinite descent. It is also a proof by contradiction, which means the proposition is proved by assuming that the opposite of the proposition is true and showing that this assumption is false, thereby implying that the proposition must be true.
Assume that √2 is a rational number, meaning that there exists an integer a and an integer b in general such that a / b = √2.
Then √2 can be written as an irreducible fraction a / b such that a and b are coprime integers and (a / b)2 = 2.
It follows that a2 / b2 = 2 and a2 = 2 b2.   ( (a / b)n = an / bn  )
Therefore a2 is even because it is equal to 2 b2. (2 b2 is necessarily even because it is 2 times another whole number and even numbers are multiples of 2.)
It follows that a must be even (as squares of odd integers are never even).
Because a is even, there exists an integer k that fulfills: a = 2k.
Substituting 2k from step 6 for a in the second equation of step 3: 2b2 = (2k)2 is equivalent to 2b2 = 4k2, which is equivalent to b2 = 2k2.
Because 2k2 is divisible by two and therefore even, and because 2k2 = b2, it follows that b2 is also even which means that b is even.
By steps 5 and 8 a and b are both even, which contradicts that a / b is irreducible as stated in step 2.
Q.E.D.

Answer 3

false, false, false. putting it over 1 does NOT make it rational.

Answer 4

irrational

Answer 5

a rational number needs whole numbers in the fraction. sqrt 2 is not a whole number

Answer 6

yes, Luis, if only sqrt(2) were an integer :D

Answer 7

Thanks everyone :)

Answer 8

pockyandsushi is correct.

don't know what infinite descent is; the classic proof is just by contradiction, which pockyandsushi gave