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Troyanec [42]
3 years ago
10

What is a counterexample of the statement “all square roots are irrational”

Mathematics
1 answer:
skad [1K]3 years ago
3 0

Rather than trying to guess and check, we can actually construct a counterexample to the statement.

So, what is an irrational number? The prefix "ir" means not, so we can say that an irrational number is something that's not a rational number, right? Since we know a rational number is a ratio between two integers, we can conclude an irrational number is a number that's not a ratio of two integers. So, an easy way to show that not all square roots are irrational would be to square a rational number then take the square root of it. Let's use three halves for our example:

\sqrt{(\frac{3}{2})^2}=\\\sqrt{\frac{9}{4}}=\\\frac{3}{2}

So clearly 9/4 is a counterexample to the statement. We can also say something stronger: All squared rational numbers are not irrational number when rooted. How would we prove this? Well, let \frac{a}{b} be a rational number. That would mean, \frac{a^2}{b^2}, would be a/b squared. Taking the square root of it yields:

\sqrt{\frac{a^2}{b^2}}}=\\ \frac{\sqrt{a^2}}{\sqrt{b^2}}=\\ \frac{a}{b}

So our stronger statement is proven, and we know that the original claim is decisively false.

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Answer:

x = 15

Step-by-step explanation:

y = family members = 7

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Substitute 7 for y:

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7x + 42 = 147

Subtract 42 on both sides:

7x + 42 = 147

     - 42    -42

7x = 105

Divide both sides by 7:

7x = 105

x = 15


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