The proof is by the method of reductio ad absurdum.
We start by assuming that sqrt(3) is rational.
That means that it can be expressed in the form p/q where p and q are co-prime integers.
Thus sqrt(3) = p/q.
This can be simplified to 3*q^2 = p^2
Now 3 divides the left hand side (LHS) so it must divide the right hand side (RHS).
That is, 3 must divide p^2 and since 3 is a prime, 3 must divide p.
That is p = 3*r for some integer r.
Then substituting for p gives,
3*q^2 = (3*r)^2 = 49*r^2
Dividing both sides by 3 gives q^2 = 3*r^2.
But now 3 divides the RHS so it must divide the LHS.
That is, 3 must divide q^2 and since 3 is a prime, 3 must divide q.
But then we have 3 dividing p as well as q which contradict the requirement that p and q are co-prime.
The contradiction implies that sqrt(3) cannot be rational.
Suppose x is a rational number and y is an irrational number.Let x + y = z, and assume that z is a rational number.The set of rational number is a group.This implies that since x is rational, -x is rational [invertibility].Then, since z and -x are rational, z - x must be rational [closure].But z - x = y which implies that y is rational.That contradicts the fact that y is an irrational number. The contradiction implies that the assumption [that z is rational] is incorrect.Thus, the sum of a rational number x and an irrational number y cannot be rational.
There is no simple method. If you can write the number as a ratio of two integers, in the form p/q where q>0, then the number is rational. But it is often very difficult to show that such a representation is not possible.
False
To demonstrate the validity of a statement using proof by absurdity or contradiction, we assume the opposite of the statement is true and then show that this assumption leads to a logical contradiction or absurdity. This contradiction proves that the original statement must be true.
False
true
Assume it's rational. Then 2 + root2 = some rational number q. Then root2 = q - 2. However, the rational numbers are well-defined under addition by (a,b) + (c,d) = (ad + bc, bd) (in other words, you can add two fractions a/b and c/d and always get another fraction of the form (ad + bc)/bd.) Therefore, q - 2 = q + (-2) is rational, since both q and -2 are rational. This implies root2 must be rational, which is a contradiction. Therefore the assumption that 2 + root2 is rational must be false.
TrueIt is true that the body of an indirect proof you must show that the assumption leads to a contradiction. In math a proof is a deductive argument for a mathematical statement.
TrueIt is true that the body of an indirect proof you must show that the assumption leads to a contradiction. In math a proof is a deductive argument for a mathematical statement.
0.4
True
It is a rational number that is 2/9 as a fraction