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Vlada [557]
3 years ago
6

Please help with question 6, 7, 8, and 9 about rational and irrational numbers

Mathematics
1 answer:
aliina [53]3 years ago
4 0
6. When we multiply two surdic integers, we can leave it into one big surd.
For instance, if we have:

\sqrt{a} \cdot \sqrt{b}, we can simplify it down to one uniform square root sign, \sqrt{ab}.

So, we can say: 3\sqrt{2} \cdot 8\sqrt{18} = 24\sqrt{36}, which is a rational root, because the square root of 36 can be converted into an integer, namely 6 and -6.

Hence, it is a rational root.

7. The summation, on the other hand, cannot be simplified any further, so it will stay as irrational. This is because √2 is already an irrational root, and adding it onto a number that is rational will stay as irrational.

8. Once again, the summation cannot be simplified. We can only simplify by factoring like terms. Hence, this must be irrational.

9. Even if it is in fractional notation, √5 can never be a rational number because no two integers when multiplied by itself will give 5. So, even if it is written in fractional form, they cannot be rational if there is an already irrational term in it.
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