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# Is the root of 7 irrational?

Updated: 12/24/2022

Wiki User

12y ago

Yes, and here's the proof:

Let's start out with the basic inequality 4 < 7 < 9.

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

2 < √7 < 3.

If you subtract all numbers by 2, you get:

0 < √7 - 2 < 1.

If √7 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 √7. Therefore, √7n must be an integer, and n must be the smallest multiple of √7 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 √7n by (√7 - 2). This gives 7n - 2√7n. Well, 7n is an integer, and, as we explained above, √7n is also an integer; therefore, 7n - 2√7n is an integer as well. We're going to rearrange this expression to (√7n - 2n)√7, and then set the term (√7n - 2n) equal to p, for simplicity. This gives us the expression √7p, which is equal to 7n - 2√7n, and is an integer.

Remember, from above, that 0 < √7 - 2 < 1.

If we multiply this inequality by n, we get 0 < √7n - 2n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √7p < √7n. We've already determined that both √7p and √7n are integers, but recall that we said n was the smallest multiple of √7 to yield an integer value. Thus, √7p < √7n is a contradiction; therefore √7 can't be rational, and so must be irrational.

Q.E.D.

Wiki User

12y ago

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