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Is the square root of 50 irrational or rational?

Updated: 10/25/2022

Wiki User

12y ago

It is irrational, here's the proof.

Let's start out with the basic inequality 49 < 50 < 64.

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

7< âˆš50 < 8.

If you subtract all numbers by 7, you get:

0 < âˆš50 - 7 < 1.

If âˆš50 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 âˆš50. Therefore, âˆš50n must be an integer, and n must be the smallest multiple of âˆš50 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 âˆš50n by (âˆš50 - 7). This gives 50n - 7âˆš50n. Well, 50n is an integer, and, as we explained above, âˆš50n is also an integer, so 7âˆš50n is an integer too; therefore, 50n - 7âˆš50n is an integer as well. We're going to rearrange this expression to (âˆš50n - 7n)âˆš50 and then set the term (âˆš50n - 7n) equal to p, for simplicity. This gives us the expression âˆš50p, which is equal to 50n - 7âˆš50n, and is an integer.

Remember, from above, that 0 < âˆš50 - 7 < 1.

If we multiply this inequality by n, we get 0 < âˆš50n - 7n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus âˆš50p < âˆš50n. We've already determined that both âˆš50p and âˆš50n are integers, but recall that we said n was the smallest multiple of âˆš50 to yield an integer value. Thus, âˆš50p < âˆš50n is a contradiction; therefore âˆš50 can't be rational and so must be irrational.

Q.E.D.

Wiki User

13y ago

Wiki User

6y ago

√50 is irrational.

Wiki User

12y ago

Irrational.

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Q: Is the square root of 50 irrational or rational?
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The square roots of 50 are irrational.

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Is the square root of 50 a rational or irrational?

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