Question

simplify. express your answer as a single term, without a denominator.
$c^0 \cdot c^{-1}$

Explanation:

Step1: Apply zero exponent rule

Any non - zero number to the power of 0 is 1. So, \(c^{0}=1\) (assuming \(c
eq0\)).

Step2: Apply product of exponents rule

When multiplying two terms with the same base \(a\), we use the rule \(a^{m}\cdot a^{n}=a^{m + n}\). Here, the base is \(c\), \(m = 0\) and \(n=- 1\). So \(c^{0}\cdot c^{-1}=c^{0+( - 1)}=c^{-1}\). But we need to express it without a denominator. Recall that \(a^{-n}=\frac{1}{a^{n}}\), so to get rid of the negative exponent, we can also note that from step 1, \(c^{0}\cdot c^{-1}=1\times c^{-1}=c^{-1}\), and another way: since we want a single term without denominator, and using the exponent addition rule directly: \(c^{0}\cdot c^{-1}=c^{0+( - 1)}=c^{-1}\), but we can rewrite \(c^{-1}\) as \(\frac{1}{c}\), but wait, no - the problem says "without a denominator". Wait, maybe I made a mistake. Wait, \(c^{0}=1\), so \(c^{0}\cdot c^{-1}=1\times c^{-1}=c^{-1}\), but to write it without a denominator, we can use the definition \(a^{-n}=\frac{1}{a^{n}}\), but we want to express it as a single term without denominator, so actually, \(c^{-1}\) is a single term, but if we consider that maybe we can write it as \(c^{-1}\) or recognize that \(c^{0}\cdot c^{-1}=c^{-1}\), and since the problem says "without a denominator", \(c^{-1}\) is a single term (it's not a fraction, it's a term with a negative exponent). Wait, maybe the intended path is:

First, use the zero - exponent property: \(c^{0}=1\) (for \(c
eq0\)). Then, use the product of powers property \(a^{m}\cdot a^{n}=a^{m + n}\). So \(c^{0}\cdot c^{-1}=c^{0+( - 1)}=c^{-1}\). But if we want to write it without a denominator, and since \(c^{-1}=\frac{1}{c}\) has a denominator, but the problem says "without a denominator", so we can leave it as \(c^{-1}\) or maybe there is a miscalculation. Wait, no, \(c^{0}\) is 1, so \(1\times c^{-1}=c^{-1}\), and \(c^{-1}\) is a single term without a denominator (it's a monomial with a negative exponent). Alternatively, maybe the problem expects us to use the exponent rules correctly:

\(c^{0}\cdot c^{-1}=c^{0+( - 1)}=c^{-1}\), and since we can write \(c^{-1}\) as \(\frac{1}{c}\), but that has a denominator. Wait, maybe I messed up the exponent addition. Wait, \(c^{0}\) is 1, so \(1\times c^{-1}=c^{-1}\), and \(c^{-1}\) is a single term (a term with a negative exponent) and does not have a denominator in the form of a fraction. So the simplified form is \(c^{-1}\) or we can also note that \(c^{0}\cdot c^{-1}=c^{-1}\), and if we want to write it as a single term without a denominator, \(c^{-1}\) is acceptable. But maybe the problem expects us to use the rule \(a^{m}\cdot a^{n}=a^{m + n}\) directly: \(c^{0}\cdot c^{-1}=c^{0+( - 1)}=c^{-1}\).

Answer:

\(c^{-1}\) (or if we consider that maybe the problem wants it in positive exponent form but without denominator, but \(c^{-1}\) is a single term without denominator. Alternatively, maybe there is a mistake in my reasoning, but following the exponent rules: \(c^{0}\cdot c^{-1}=c^{-1}\))