Radicals

A counterexample is one specific choice of numbers that shows a supposed rule does not always work. To disprove $\sqrt{a-b}=\sqrt{a}-\sqrt{b}$ as a general rule, we must find values of $a$ and $b$ where the left-hand side (LHS) and right-hand side (RHS) give different results.

Use the numbers from Option A: $a=81$ and $b=49$. Compute the subtraction inside the radical first:

$\begin{array}{rl}\text{LHS}& =\sqrt{81-49}\\ & =\sqrt{32}\end{array}$

Since $\sqrt{32}$ is not equal to $2$ (and is about $5.66$), we keep it as $\sqrt{32}$ for an exact comparison later.

Now take the square roots separately, then subtract:

$\begin{array}{rl}\text{RHS}& =\sqrt{81}-\sqrt{49}\\ & =9-7\\ & =2\end{array}$

From the calculations, the LHS is $\sqrt{32}$ and the RHS is $2$. Because $\sqrt{32}\ne 2$, this shows the equation $\sqrt{a-b}=\sqrt{a}-\sqrt{b}$ fails for these values. Therefore, Option A is a valid counterexample.

Here is why the other options are not suitable counterexamples for the subtraction rule:

• Option B uses a plus sign ($+$), so it is testing a different (addition) idea rather than the subtraction statement.
• Option C sets $b=0$. That can make the expression true in that special case, but a counterexample must show a situation where the rule breaks.
• Option D is about multiplication, and it shows a true radical property ($\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$), so it does not help disprove the subtraction rule.

The statement that correctly acts as a counterexample is Option A.

We need a counterexample to the claim:

$\sqrt{a-b}=\sqrt{a}-\sqrt{b}$

A counterexample must use subtraction and must make the two sides unequal. We can eliminate options that test other operations or that do not actually disprove anything.

Option B involves addition:

$\sqrt{36+25}\ne \sqrt{36}+\sqrt{25}$

Even though it’s true that $\sqrt{61}\ne 11$, this would be relevant to a different (and also false) statement: $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$. It does not address the subtraction rule we are trying to disprove.

Option D uses multiplication:

$\sqrt{81\cdot 49}=\sqrt{81}\cdot \sqrt{49}$

This matches a correct radical rule. In fact, $\sqrt{81\cdot 49}=\sqrt{3969}=63$ and $\sqrt{81}\cdot \sqrt{49}=9\cdot 7=63$, so it cannot be used to show the subtraction rule is false.

Option C is:

$\sqrt{49-0}=\sqrt{49}-\sqrt{0}$

This is true because both sides equal $7$. But one example where the rule works does not prove it works for all numbers. To disprove a general rule, we need an example where it fails.

Option A tests subtraction with non-zero values:

$\begin{array}{rl}\text{Left Side: }& \sqrt{81-49}=\sqrt{32}\\ \text{Right Side: }& \sqrt{81}-\sqrt{49}=9-7=2\end{array}$

Because $\sqrt{32}\ne 2$, this option successfully shows the statement is not true in general. So Option A is the correct counterexample.

Using the Combining Radicals idea, we rely on the Product Property of square roots. It tells us that multiplying two square roots is the same as taking the square root of the product inside:

$\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$

This rule is a key skill in the Radicals unit.

Use the property on the given expression by multiplying the radicands (the values inside the roots):

$\sqrt{2}\cdot \sqrt{8}=\sqrt{2\cdot 8}$

$\sqrt{2\cdot 8}=\sqrt{16}$

Next simplify $\sqrt{16}$. Because $16$ is a perfect square ($4^2=16$), its square root is a whole number:

$\sqrt{16}=4$

The value $4$ is an integer, and any integer can be written as a fraction (for example, $\frac{4}{1}$). That means the answer is rational.

Remember: irrational numbers have decimals that do not terminate and do not repeat (like $\sqrt{2}$), while rational numbers can be expressed as a ratio of integers.

Check why the other choices are not correct:

• Option A is wrong because the simplified value is not $2\sqrt{2}$.
• Option C is wrong because multiplying two surds can sometimes give a rational result (as it does here).
• Option D is wrong because it incorrectly says $\sqrt{2}$ and $\sqrt{8}$ are rational; both are actually irrational.

Since the product simplifies exactly to $4$, the result is rational. That corresponds to Option B.

With the Simplifying Individual Surds method, start by simplifying $\sqrt{8}$. Find the greatest perfect-square factor of 8.

Because $8=4\times 2$ and 4 is a perfect square, we can write:

$\sqrt{8}=\sqrt{4\cdot 2}=\sqrt{4}\cdot \sqrt{2}=2\sqrt{2}$

Replace $\sqrt{8}$ with $2\sqrt{2}$ in the original product:

$\sqrt{2}\cdot 2\sqrt{2}$

Reorder the factors (multiplication can be rearranged):

$2\cdot (\sqrt{2}\cdot \sqrt{2})$

And since $\sqrt{a}\cdot \sqrt{a}=a$:

$2\cdot 2=4$

Now that we have the value 4, compare against the statements:

• Option A is incorrect because the answer is 4, not $2\sqrt{2}$.
• Option C is incorrect because even though $\sqrt{2}$ and $\sqrt{8}$ are surds, their product here is rational.
• Option D is incorrect because $\sqrt{2}$ and $\sqrt{8}$ are not rational numbers.

Because $4$ can be written as $\frac{4}{1}$, it is rational, matching the remaining choice.

Start by evaluating the multiplication. Using $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$:

$\sqrt{2}\cdot \sqrt{8}=\sqrt{2\cdot 8}=\sqrt{16}$

And since $4^2=16$:

$\sqrt{16}=4$

Option D says $\sqrt{2}$ and $\sqrt{8}$ are rational. But surds (square roots of non-perfect squares) are irrational, so this claim is not true.

Option A states the expression becomes $2\sqrt{2}$. Our calculation shows the result is $4$, and $4\ne 2\sqrt{2}$. Also, $4$ is not irrational.

Option C argues the product must be irrational because it involves surds. However, $\sqrt{2}\cdot \sqrt{8}=\sqrt{16}=4$, which is rational. Multiplying irrational numbers does not always keep the result irrational (for example, $\sqrt{2}\cdot \sqrt{2}=2$).

After eliminating the others, Option B matches: the expression simplifies to $4$, and since $4=\frac{4}{1}$, it is a rational number.

The claim uses the word any, meaning it says every pair of irrational numbers will add to an irrational number. To show a universal statement is wrong, we can use the Counterexample Method: find just one pair of irrational numbers whose sum is actually rational, and the whole claim is disproved.

Pick two numbers that are both irrational.

A well-known irrational number is $\sqrt{2}$.

Its opposite, $-\sqrt{2}$, is also irrational.

Each is irrational because it cannot be written in the form $\frac{p}{q}$ where $p$ and $q$ are integers (this connects to Topic: The number system).

Now add them:

$\sqrt{2}+(-\sqrt{2})=0$

The sum is $0$.

Is $0$ irrational? No.

$0$ is an integer, and every integer is rational because it can be written as a fraction, such as $\frac{0}{1}$.

Because we found two irrational numbers that add to a rational number ($0$), the statement that the sum is always irrational cannot be correct.

One counterexample is enough to disprove the statement, so it is false.

A set is closed under an operation (like addition) if doing that operation to any two numbers in the set always gives a number that is still in the same set.

So to judge the claim, we are really checking whether irrational numbers are closed under addition, a key idea in Topic: The number system.

The statement says that adding any two irrational numbers will always give an irrational number.

If we can find even one example where two irrational numbers add to a rational number, then irrational numbers are not closed under addition, and the statement must be false.

Choose two irrational numbers:

1. $a=\sqrt{2}$
2. $b=-\sqrt{2}$

Both are irrational since they cannot be written as a ratio of integers. Add them:

$\sqrt{2}+(-\sqrt{2})=0$

But $0$ is an integer, so it is rational (for instance, $\frac{0}{1}$).

Since two irrational numbers can add to a rational number ($0$), the result is not guaranteed to stay irrational. That means irrational numbers are not closed under addition, so the statement cannot be correct.

Because the irrational numbers are not closed under addition, the correct choice is False.

The claim says the sum of any two irrational numbers is always irrational. To disprove it, we only need one situation where the sum turns out rational.

This links to The number system: irrational numbers cannot be written as $\frac{p}{q}$ with integers $p$ and $q$.

A quick way to create a rational result from irrationals is to use algebraic conjugates, which look like:

$a+\sqrt{b}\text{and}a-\sqrt{b}$

Take the simple case where $a=0$ and $\sqrt{b}=\sqrt{2}$. Then:

1. $x=\sqrt{2}$
2. $y=-\sqrt{2}$

Both are irrational because $\sqrt{2}$ cannot be written as a simple fraction.

Compute the sum:

$\text{Sum}=\sqrt{2}+(-\sqrt{2})$

$\text{Sum}=0$

And since $0=\frac{0}{1}$, the sum is rational. So we have two irrational numbers whose sum is rational.

Consider another conjugate pair: $3+\sqrt{5}$ and $3-\sqrt{5}$.

Each is irrational, but adding them gives:

$(3+\sqrt{5})+(3-\sqrt{5})=3+3+\sqrt{5}-\sqrt{5}=6$

And $6$ is rational. This supports that the sum of two irrationals is not guaranteed to be irrational.

Because we have clear examples where the sum becomes rational, the statement cannot hold for all cases.

Since the claim fails in at least one case, the correct answer is False.