Powers, roots and exponents

To simplify the expression $(-1{)}^n\times (-1{)}^{n+1}$, observe that both factors share the same base, $-1$. When we multiply powers that have an identical base, we use the Product Rule for Exponents, a key idea in Topic 1: Number and Algebra Foundations in the IB MYP syllabus.

The rule states:

$a^m\times a^n=a^{m+n}$

Using the product rule, we keep the base $-1$ and add the exponents $n$ and $n+1$:

$\begin{array}{rl}(-1{)}^n\times (-1{)}^{n+1}& =(-1{)}^{n+(n+1)}\\ & =(-1{)}^{2n+1}\end{array}$

Next, look at the exponent $2n+1$ for any integer $n$:

• $2n$ is always even because it is divisible by 2.
• Adding 1 to an even number makes it odd, so $2n+1$ is always odd.

Remember: $(-1)$ to an odd power is $-1$, and $(-1)$ to an even power is $1$.

Since $2n+1$ is always odd:

$(-1{)}^{2n+1}=-1$

Compare our result to the answer choices:

• Option A says 0, but powers of $-1$ never produce 0.
• Option B is $(-1{)}^{2n}$. Because $2n$ is even, this equals $1$, not our result.
• Option C is $1$, but our exponent $2n+1$ is always odd, so the value cannot be $1$.

So, for every integer $n$, the expression always simplifies to $-1$.

Because the expression must simplify the same way for any integer $n$, we can plug in a few easy integers (like $n=0$ and $n=1$) and see which option consistently matches.

Replace $n$ with 0:

$\begin{array}{rl}& (-1{)}^0\times (-1{)}^{0+1}\\ & =1\times (-1{)}^1\\ & =1\times (-1)\\ & =-1\end{array}$

So when $n=0$, the value is $-1$.

Now try $n=1$:

$\begin{array}{rl}& (-1{)}^1\times (-1{)}^{1+1}\\ & =(-1)\times (-1{)}^2\\ & =(-1)\times 1\\ & =-1\end{array}$

Both checks give $-1$.

Match $-1$ against the choices:

• Option A: gives $0$, so it doesn’t work.
• Option C: gives $1$, so it doesn’t work.
• Option B: if $n=1$, then $(-1{)}^{2(1)}=(-1{)}^2=1$, which doesn’t match $-1$.

Since substitution keeps giving $-1$, Option D is the only answer that works for every integer $n$.

We want to simplify $(-1{)}^n\times (-1{)}^{n+1}$.

The exponents are $n$ and $n+1$, which are consecutive integers. Using Consecutive Parity Analysis, we think about whether each exponent is even or odd.

Since $n$ and $n+1$ are consecutive, one must be even and the other must be odd:

$\begin{array}{rl}& \text{Case 1: }n\text{ is even}⟹n+1\text{ is odd}\\ & \text{Case 2: }n\text{ is odd}⟹n+1\text{ is even}\end{array}$

From exponent patterns in Number and Algebra Foundations:

• If $k$ is even, $(-1{)}^k=1$
• If $k$ is odd, $(-1{)}^k=-1$

Now compute the product in both situations:

$\begin{array}{rl}& \text{Case 1: }(1)\times (-1)=-1\\ & \text{Case 2: }(-1)\times (1)=-1\end{array}$

No matter whether $n$ is even or odd, the product is always $-1$.

Since the value is always $-1$, the other options must be incorrect:

• Option A is incorrect because the product is not 0.
• Option B, $(-1{)}^{2n}$, equals $((-1{)}^2{)}^n=1^n=1$, so it cannot be right.
• Option C is incorrect because the result is negative, not positive.

Therefore, the product of $(-1)$ raised to two consecutive integers is always $-1$.

We start by remembering the quotient rule from the exponent laws. When you divide powers that have the same non-zero base $a$, you subtract the exponents:

$\frac{a^m}{a^n}=a^{m-n}$

This rule is only used when the base is the same and $a\ne 0$.

To make a zero exponent appear, choose a case where the top and bottom exponents are equal (take $n=m$).

Then the quotient rule gives:

$\frac{a^m}{a^m}=a^{m-m}=a^0$

Now evaluate the exact same expression using ordinary arithmetic. Any non-zero quantity divided by itself equals $1$:

$\frac{a^m}{a^m}=1$

Since one viewpoint gives $a^0$ and the other gives $1$ for the same fraction, it follows that:

$a^0=1$

We can reject the other options because they do not use the quotient rule to derive $a^0=1$:

• Option A is based on the product rule ($a^m\times a^n=a^{m+n}$), not division.
• Option C is about the power of a power rule ($(a^m{)}^n=a^{m\times n}$).
• Option D includes $a^0$ in a denominator but does not explain where $a^0=1$ comes from; it treats it as already true.

Option B is exactly the quotient-rule setup we used: it shows the exponents being subtracted so that $m-m$ becomes $0$, giving $a^0$.

First, identify the Quotient Rule for indices. It tells you how to divide powers with the same base by subtracting exponents:

$\frac{a^m}{a^n}=a^{m-n}$

We need the option that uses this rule to justify why $a^0=1$.

Check Option A: $a^m\times a^{-m}=a^{m+(-m)}=a^0$.

This is using the Product Rule ($a^m\cdot a^n=a^{m+n}$) because it multiplies, not divides. So it cannot be the correct quotient-rule identity.

Now look at Option C: $(a^1{)}^0=a^0$.

This connects to the Power of a Power Rule, $(a^m{)}^n=a^{mn}$. Since the question asks specifically for the quotient rule, we eliminate this choice.

Consider Option D: $a^n÷a^0=a^n$.

Even though it involves division, it does not prove anything about $a^0$. It only works if you already accept that dividing by $a^0$ changes nothing—so it assumes the result instead of deriving it by subtracting exponents.

A calculator check supports the idea: pick a non-zero number such as $5$ and divide it by itself. Basic arithmetic gives $1$, and the quotient rule predicts an exponent difference of $0$, so both viewpoints agree.

After eliminating the others, Option B remains:

$\frac{a^m}{a^m}=a^{m-m}=a^0$

This correctly applies the Quotient Rule because it subtracts the exponents ($m-m=0$). Since $\frac{a^m}{a^m}=1$ for $a\ne 0$, this identity demonstrates that $a^0=1$.