Writing and Proof: If true prove it, if false give a counterexample. Use contradiction when proving. (a) For each positive real number x, if x is irrational, then x^2 is irrational. (b) For every pair of real numbers a nd y, if x+y is irrational, then x if irrational or y is irrational

Writing and Proof: If true prove it, if false give a counterexample. Use contradiction when proving. (a) For each positive real number x, if x is irrational, then x^2 is irrational. (b) For every pair of real numbers a nd y, if x+y is irrational, then x if irrational or y is irrational

Question
Irrational numbers
asked 2020-12-14
Writing and Proof: If true prove it, if false give a counterexample. Use contradiction when proving.
(a) For each positive real number x, if x is irrational, then \(\displaystyle{x}^{{2}}\) is irrational.
(b) For every pair of real numbers a nd y, if x+y is irrational, then x if irrational or y is irrational

Answers (1)

2020-12-15
(a) The given statement is false. Because for each positive real number x, if x is irrational number, then it is not necessary to their square \(\displaystyle{x}^{{2}}\) is also irrational. For example:
consider \(\displaystyle{x}=\sqrt{{5}}\) ((it is positive real number) Then,
\(\displaystyle{x}^{{2}}={\left(\sqrt{{5}}\right)}^{{2}}\)
\(\displaystyle{x}^{{2}}={5}\) {since, \(\displaystyle{\left(\sqrt{{a}}\right)}^{{2}}={a}\)}
Since, 5 is not an irrational number because it can be written in the form of \(\displaystyle\frac{{p}}{{q}}\) {where p and q are integer}.
That is \(\displaystyle{5}=\frac{{5}}{{1}}\).
Hence, given statement is false.
(b)The given statement is true.To prove this, contrary assume that x and y both are rational number. So, \(\displaystyle{x}=\frac{{a}}{{b}}\) {where a and b are the integers}
\(\displaystyle{y}=\frac{{c}}{{d}}\) {where c and d are the integers}
According to the given statement:
x+y=irrational number
\(\displaystyle\frac{{a}}{{b}}+\frac{{c}}{{d}}\)=ZSK irrational number
\(\displaystyle\frac{{{a}{d}+{c}{d}}}{{{b}{d}}}=\) irrational number
Let, ad+cd=p and bd=q.
Then,
\(\displaystyle\frac{{p}}{{q}}=\) irrational number
rational number=irrational number
Which is the contradiction. Therefore, our consideration values {x and y both are rational number} are wrong. Hence, the given statement is true.
For example:
\(\displaystyle{x}={2},{y}=\sqrt{{3}}\), then \(\displaystyle{x}+{y}={2}+\sqrt{{3}}=\) irrational
0

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A random sample of \( n_1 = 14 \) winter days in Denver gave a sample mean pollution index \( x_1 = 43 \).
Previous studies show that \( \sigma_1 = 19 \).
For Englewood (a suburb of Denver), a random sample of \( n_2 = 12 \) winter days gave a sample mean pollution index of \( x_2 = 37 \).
Previous studies show that \( \sigma_2 = 13 \).
Assume the pollution index is normally distributed in both Englewood and Denver.
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\( H_0:\mu_1=\mu_2.\mu_1>\mu_2 \)
\( H_0:\mu_1<\mu_2.\mu_1=\mu_2 \)
\( H_0:\mu_1=\mu_2.\mu_1<\mu_2 \)
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The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
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At the \( \alpha = 0.01 \) level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the \( \alpha = 0.01 \) level, we reject the null hypothesis and conclude the data are statistically significant.
At the \( \alpha = 0.01 \) level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the \( \alpha = 0.01 \) level, we reject the null hypothesis and conclude the data are not statistically significant.
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Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
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Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
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