Is the square root of 14 an irrational number?

Answer 1

Yes, here's the proof.

Let's start out with the basic inequality 81 < 83 < 100.

Now, we'll take the square root of this inequality:

9 < √83 < 10.

If you subtract all numbers by 9, you get:

0 < √83 - 9 < 1.

If √83 is rational, then it can be expressed as a fraction of two integers, m/n. This next part is the only remotely tricky part of this proof, so pay attention. We're going to assume that m/n is in its most reduced form; i.e., that the value for n is the smallest it can be and still be able to represent √83. Therefore, √83n must be an integer, and n must be the smallest multiple of √83 to make this true. If you don't understand this part, read it again, because this is the heart of the proof.

Now, we're going to multiply √83n by (√83 - 9). This gives 83n - 9√83n. Well, 83n is an integer, and, as we explained above, √83n is also an integer, so 9√83n is an integer too; therefore, 83n - 9√83n is an integer as well. We're going to rearrange this expression to (√83n - 9n)√83 and then set the term (√83n - 9n) equal to p, for simplicity. This gives us the expression √83p, which is equal to 83n - 9√83n, and is an integer.

Remember, from above, that 0 < √83 - 9 < 1.

If we multiply this inequality by n, we get 0 < √83n - 9n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √83p < √83n. We've already determined that both √83p and √83n are integers, but recall that we said n was the smallest multiple of √83 to yield an integer value. Thus, √83p < √83n is a contradiction; therefore √83 can't be rational and so must be irrational.

Q.E.D.

Answer 2

Yes, here's the proof.

Let's start out with the basic inequality 9 < 14 < 16.

Now, we'll take the square root of this inequality:

3 < √14 < 4.

If you subtract all numbers by 3, you get:

0 < √14 - 3 < 1.

If √14 is rational, then it can be expressed as a fraction of two integers, m/n. This next part is the only remotely tricky part of this proof, so pay attention. We're going to assume that m/n is in its most reduced form; i.e., that the value for n is the smallest it can be and still be able to represent √14. Therefore, √14n must be an integer, and n must be the smallest multiple of √14 to make this true. If you don't understand this part, read it again, because this is the heart of the proof.

Now, we're going to multiply √14n by (√14 - 3). This gives 14n - 3√14n. Well, 14n is an integer, and, as we explained above, √14n is also an integer, so 3√14n is an integer too; therefore, 14n - 3√14n is an integer as well. We're going to rearrange this expression to (√14n - 3n)√14 and then set the term (√14n - 3n) equal to p, for simplicity. This gives us the expression √14p, which is equal to 14n - 3√14n, and is an integer.

Remember, from above, that 0 < √14 - 3 < 1.

If we multiply this inequality by n, we get 0 < √14n - 3n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √14p < √14n. We've already determined that both √14p and √14n are integers, but recall that we said n was the smallest multiple of √14 to yield an integer value. Thus, √14p < √14n is a contradiction; therefore √14 can't be rational and so must be irrational.

Q.E.D.

Answer 3

no

Answer 4

yes