Is the square root of 50 a irrational number?

Answer 1

7.071678118.......

rounded to

7.1

*****

To a whole lot of places, it's actually 7.0710678118654752440084436210485...

But no finite number of places can be enough to establish its irrationality.

Certainly, √50 is indeed irrational. Here is how we can see that it is so:

First of all, we note that √50 = 5√2.

To show that the above is irrational, we need to show two things:

(1) that √2 is irrational; and

(2) that the product of a integer and an irrational is irrational.

We show (1), above, by reductio ad absurdum: We assume the opposite of that which we wish to prove and show that it entails a contradiction.

Suppose that √2 is rational. Then it must equal the quotient p/q, where p and q are two integers that are relatively prime. In particular, not both, p and q may be even numbers.

(√2 might also equal some p'/q', where p' and q' have a common divisor greater than 1; but, in that case, the fraction could be reduced to p/q. Thus, we can assume, without loss of generality, that p and q are relatively prime; that is, that p/q is irreducible.)

Before continuing our proof, we shall need to establish the following: The square of every even integer is an even integer; and the square of every odd integer is an odd integer.

Let n be even; thus, n = 2k, where k is some integer. Then, n2 = 4k2; thus, n2 is divisible by 4 and, therefore, it is also divisible by 2.

Let m be odd; thus, m = 2j + 1.

Then m2 = (2j + 1)2 = 4j2 + 4j + 1 = 2(2j2 + 2j) + 1, which is evidently odd.

We may conclude, from the above, that, if a perfect square is an even integer, then its square root must also be even. Now, we may continue the main proof:

From √2 = p/q, which we assumed above, p and q being relatively prime, we deduce the following:

p2 = 2q2; therefore, p2 is even; therefore, p is even.

Because p is even, p = 2r, for some integer r. Then, p2 = 4r2 = 2q2; thus, q2 = 2r2, and q is similarly shown to be even. P and q, therefore can not be relatively prime.

But, wait! Our assumption was that p and q are relatively prime and can not both be even; and, so, we have the contradiction we sought.

Therefore, (1) is proved, and there exist no integers p and q, such that p/q = √2, and p/q is an irreducible fraction. In other words, √2 must, after all, be irrational.

Now, we proceed to show (2), that the product of an integer and an irrational is irrational; it will suffice to do this by means of the example in hand: namely, that, if √2 is irrational, then √50 = 5√2 must also be irrational.

First, we know that the sum of two integers is also an integer; likewise, by induction, the product of two integers is an integer. From this, we can show that the product of two rationals is also a rational:

Let the two rationals be a/b and c/d, where a, b, c, and d are all integers. Then, their product is ac/bd, and, because ac and bd are integers, ac/bd must, by definition, be rational.

Multiplying both sides by 1/5, we obtain (1/5)(√50) = √2.

Now, 1/5 is rational; then, if √50 is rational, then √2, being the product of two rationals, must also be rational. But we have shown, in (1) above, that √2 is not rational.

We conclude that √50 can not be rational, either; therefore, it is irrational, which is what we set out to prove.