Question

priscilla investigates the values of $(5^r)(5^s)$ and $(5^r)^s$ where $r$ and $s$ are nonzero integers. she makes the following claims. - claim 1: when $r$ is a negative integer and $s$ is a positive integer, the value of $(5^r)(5^s)$ can never be a whole number. for example, the value of $(5^{-5})(5^3)$ is $0.04$. - claim 2: when $r$ is a negative integer and $s$ is a positive integer, the value of $(5^r)^s$ can never be a whole number. for example, the value of $(5^{-4})^1$ is $0.0016$. which statement about priscillas claims is true? a. both claims are correct because the provided example proves each associated claim. b. both claims are incorrect because the provided example is not sufficient to prove each associated claim.

Explanation:

To determine the validity of Priscilla's claims, we analyze each claim:

Analyzing Claim 1:

The expression \((5^r)(5^s)\) can be simplified using the exponent rule \(a^m \cdot a^n = a^{m + n}\). So, \((5^r)(5^s)=5^{r + s}\). If \(r\) is negative and \(s\) is positive, let's take an example where \(r=-3\) and \(s = 5\). Then \(r + s=-3 + 5 = 2\), and \(5^{2}=25\), which is a whole number. The example given by Priscilla (\(r=-5\), \(s = 3\), \(r + s=-2\), \(5^{-2}=\frac{1}{25}=0.04\)) is just one case, but it does not prove the general claim that \((5^r)(5^s)\) can never be a whole number.

Analyzing Claim 2:

The expression \((5^r)^s\) can be simplified using the exponent rule \((a^m)^n=a^{m\times n}\). So, \((5^r)^s = 5^{r\times s}\). If \(r\) is negative and \(s\) is positive, let's take an example where \(r = -2\) and \(s=2\). Then \(r\times s=- 4\)? Wait, no, \(r=-2\), \(s = 2\), \(r\times s=-4\)? Wait, no, \(r=-1\) and \(s = 2\), then \(r\times s=-2\), \(5^{-2}=\frac{1}{25}\) is not a whole number. Wait, another example: \(r=-2\) and \(s = 0\), but \(s\) must be positive. Wait, let's take \(r=-1\) and \(s = 2\), \(5^{-2}=\frac{1}{25}\). But if \(r=-2\) and \(s = 1\), \(5^{-2}=\frac{1}{25}\). Wait, is there a case where \((5^r)^s\) is a whole number? Let's see, if \(r=-1\) and \(s = 0\), but \(s\) is positive. Wait, maybe my first approach was wrong. Wait, the general form is \(5^{r\times s}\). For \(5^{r\times s}\) to be a whole number, \(r\times s\) must be a non - negative integer. If \(r\) is negative and \(s\) is positive, \(r\times s\) is negative. So \(5^{r\times s}=\frac{1}{5^{\vert r\times s\vert}}\), which is a fraction with denominator a power of 5, so it will be a decimal that is not a whole number? Wait, no, wait \(5^{0}=1\) is a whole number, but \(s\) is positive, so \(r\times s\) can never be 0 (since \(r
eq0\) and \(s
eq0\)). Wait, but the example given by Priscilla: \((5^{-4})^1=5^{-4}=\frac{1}{625}=0.0016\). But let's check the general claim. The claim is "can never be a whole number". But is there a case where it can be? Wait, if \(r\) is negative and \(s\) is positive, \(r\times s\) is negative, so \(5^{r\times s}=\frac{1}{5^{\vert r\times s\vert}}\), which is a positive fraction less than 1 (since \(\vert r\times s\vert>0\)), so it can never be a whole number? Wait, but the first claim: \((5^r)(5^s)=5^{r + s}\). If \(r=-2\) and \(s = 3\), \(r + s = 1\), \(5^{1}=5\) which is a whole number. So the example given by Priscilla for claim 1 is not sufficient to prove the general claim, and the example for claim 2: let's see, is there a case where \((5^r)^s\) is a whole number? If \(r=-1\) and \(s = 0\), but \(s\) is positive. So maybe claim 2 is correct? Wait, no, my initial analysis for claim 1 was correct. The key point is that a single example cannot prove a "never" claim. A "never" claim requires a general proof, not just one example. So both claims are incorrect because the provided examples are not sufficient to prove the general claims (since we can find counter - examples for claim 1, and for claim 2, maybe the general case holds, but the example alone does not prove the "never" claim).

Answer:

B. Both claims are incorrect because the provided example is not sufficient to prove each associated claim.