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chubhunter [2.5K]
2 years ago
11

Explain why any of the four operations placed between two terms 5 and -3 sqrt (8) will result in an irrational number

Mathematics
1 answer:
Gnoma [55]2 years ago
7 0

Sum/difference:

Let

x = 5 + (-3\sqrt{8}) = 5-3\sqrt{8}

This means that

3\sqrt{8} = 5-x \iff \sqrt{8} = \dfrac{5-x}{3}

Now, assume that x is rational. The sum/difference of two rational numbers is still rational (so 5-x is rational), and the division by 3 doesn't change this. So, you have that the square root of 8 equals a rational number, which is false. The mistake must have been supposing that x was rational, which proves that the sum/difference of the two given terms was irrational

Multiplication/division:

The logic is actually the same: if we multiply the two terms we get

x = -15\sqrt{8}

if again we assume x to be rational, we have

\sqrt{8} = -\dfrac{x}{15}

But if x is rational, so is -x/15, and again we come to a contradiction: we have the square root of 8 on one side, which is irrational, and -x/15 on the other, which is rational. So, again, x must have been irrational. You can prove the same claim for the division in a totally similar fashion.

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