Real Numbers - Class 10 Mathematics - Chapter 1 - Notes, NCERT Solutions & Extra Questions
Renews every month. Cancel anytime
Your personal doubt-solving assistant
Chatterbot AI gives you 100% accurate answers to your questions in an instant.
Exercise 1.1 - Real Numbers | NCERT | Mathematics | Class 10
Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Here are the prime factorizations of the given numbers:
(i) $140 = 2^2 \times 5 \times 7$
(ii) $156 = 2^2 \times 3 \times 13$
(iii) $3825 = 3^2 \times 5^2 \times 17$
(iv) $5005 = 5 \times 7 \times 11 \times 13$
(v) $7429 = 17 \times 19 \times 23$
Find the LCM and HCF of the following pairs of integers and verify that $\mathrm{LCM} \times \mathrm{HCF}=$ product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Improve your grades!
Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.
Sign up now![Boost your exam scores](https://www.englishchatterbox.com/frontend/v4/assets/img/rocket-launch.png)
Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8,9 and 25
Improve your grades!
Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.
Sign up now![Boost your exam scores](https://www.englishchatterbox.com/frontend/v4/assets/img/rocket-launch.png)
Given that $\operatorname{HCF}(306,657)=9$, find $\operatorname{LCM}(306,657)$.
Improve your grades!
Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.
Sign up now![Boost your exam scores](https://www.englishchatterbox.com/frontend/v4/assets/img/rocket-launch.png)
Check whether $6^{n}$ can end with the digit 0 for any natural number $n$.
Improve your grades!
Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.
Sign up now![Boost your exam scores](https://www.englishchatterbox.com/frontend/v4/assets/img/rocket-launch.png)
Explain why $7 \times 11 \times 13+13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1+5$ are composite numbers.
Improve your grades!
Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.
Sign up now![Boost your exam scores](https://www.englishchatterbox.com/frontend/v4/assets/img/rocket-launch.png)
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Improve your grades!
Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.
Sign up now![Boost your exam scores](https://www.englishchatterbox.com/frontend/v4/assets/img/rocket-launch.png)
💡 Have more questions?
Ask Chatterbot AIExercise 1.2 - Real Numbers | NCERT | Mathematics | Class 10
Prove that $\sqrt{5}$ is irrational.
To prove that (\sqrt{5}) is irrational, we'll use a proof by contradiction. Here are the steps:
-
Assume the opposite: Assume that (\sqrt{5}) is rational. This means it can be written as a fraction (\frac{a}{b}), where (a) and (b) are coprime integers (they have no common positive divisor other than 1), and (b \neq 0).
-
Expressing the Assumption: Based on the assumption, (\sqrt{5} = \frac{a}{b}). Squaring both sides gives us (5 = \frac{a^2}{b^2}), which implies (a^2 = 5b^2).
-
Contradiction on Integrality: The last equation implies that (a^2) is divisible by 5. Therefore, (a) must also be divisible by 5 (because the square of a non-multiple of 5 is never a multiple of 5). Let (a = 5k) for some integer (k).
-
Substituting Back: Substitute (a = 5k) back into (a^2 = 5b^2). We get ((5k)^2 = 5b^2), which simplifies to (25k^2 = 5b^2), and further to (5k^2 = b^2).
-
Contradiction on (b): The equation (5k^2 = b^2) implies that (b^2), and hence (b), is divisible by 5. This is a contradiction because we initially assumed that (a) and (b) are coprime, meaning they cannot both be divisible by 5.
-
Conclusion: Since assuming that (\sqrt{5}) is rational leads to a contradiction, (\sqrt{5}) must be irrational.
This proof by contradiction confirms the irrationality of (\sqrt{5}).
Prove that $3+2 \sqrt{5}$ is irrational.
Improve your grades!
Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.
Sign up now![Boost your exam scores](https://www.englishchatterbox.com/frontend/v4/assets/img/rocket-launch.png)
Prove that the following are irrationals :
(i) $\frac{1}{\sqrt{2}}$
(ii) $7 \sqrt{5}$
(iii) $6+\sqrt{2}$
Improve your grades!
Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.
Sign up now![Boost your exam scores](https://www.englishchatterbox.com/frontend/v4/assets/img/rocket-launch.png)
💡 Have more questions?
Ask Chatterbot AIExtra Questions - Real Numbers | NCERT | Mathematics | Class 10
For any integers $a$ and $3$, there exists unique integers $q$ and $r$ such that $a = 3q + r$. Find the possible value of $r$.
By applying Euclid's division lemma, which states that for any two integers $ a $ and $ b $, there exist unique integers $ q $ (the quotient) and $ r $ (the remainder) such that:
$$ a = bq + r $$
and
$$ 0 \leq r < b $$
For the given question, we can substitute $ b = 3 $ into the lemma:
$$ a = 3q + r $$
Based on the condition provided by Euclid's division lemma, the value of $ r $ must satisfy:
$$ 0 \leq r < 3 $$
Therefore, the possible values of $ r $ are:
$0$
$1$
$2$
The number of positive integers satisfying the equation $x + \log_{10}\left(2^{x} + 1\right) = x \log_{10} 5 + \log_{10} 6$ is
A) 0
B) 1
C) 2
D) infinite
The correct option is B) 1.
Here is the step-by-step solution:
Transform the original equation: $$ x + \log_{10}(2^x + 1) = x \log_{10} 5 + \log_{10} 6 $$
Rewrite it using logarithmic properties: $$ \log_{10}10^x + \log_{10}(2^x + 1) = \log_{10}5^x + \log_{10}6 $$
Combine the terms under a single logarithm using the product rule for logs: $$ \log_{10}\left(10^x (2^x + 1)\right) = \log_{10}(5^x \cdot 6) $$
Since the logarithms are equal, the arguments must be equal: $$ 10^x(2^x + 1) = 5^x \cdot 6 $$
Simplify assuming that $10^x = 5^x \cdot 2^x$: $$ 2^x(2^x + 1) = 6 \quad (\because 5^x \neq 0) $$
This yields a quadratic equation in terms of $2^x$: $$ (2^x)^2 + 2^x - 6 = 0 $$
Solve the quadratic equation: $$ 2^x = 2 \text{ or } 2^x = -3 \quad (\text{-3 is not possible}) $$
Only $2^x = 2$ is valid since $2^x$ cannot be negative. This leads to: $$ x = 1 $$
Therefore, there is one solution, $x = 1$.
Steve is standing at number 38 on a number line. If he jumps 7 steps to the right and then again 3 steps to the right, where will he be standing now?
A) 41
B) 45
C) 48
D) He will fall off the number line.
Steve initially stands at number 38 on the number line. To determine his new position after a sequence of jumps to the right, we follow these calculations:
Starting at 38, Steve takes 7 steps to the right. On a number line, moving to the right means we add the steps to the current number: $$ 38 + 7 = 45 $$
Next, he takes another 3 steps to the right. Continuing from 45, we add these 3 steps: $$ 45 + 3 = 48 $$
Therefore, after jumping a total of 10 steps to the right, Steve will be standing at number 48. This concludes that the correct option is:
C) 48
The number of real solutions of the equation $\cos \left(e^{x}\right) = 2^{x} + 2^{-x}$ is (a) 0 (b) 1 (c) 2 (d) infinitely many
To determine the number of real solutions for the equation: $$ \cos(e^x) = 2^x + 2^{-x} $$
First, analyze the function on the right-hand side, $2^x + 2^{-x}$. This function can be understood by considering that $2^x$ and $2^{-x}$ are always positive for any real $x$, due to the exponential nature of the terms.
By applying the Arithmetic Mean - Geometric Mean Inequality (AM-GM Inequality), we get: $$ \frac{2^x + 2^{-x}}{2} \geq \sqrt{2^x \cdot 2^{-x}} $$ Simplifying the terms, $$ \frac{2^x + 2^{-x}}{2} \geq \sqrt{1} = 1 $$ Thus, multiplying through by 2, we find: $$ 2^x + 2^{-x} \geq 2 $$
Next, consider the left-hand side of the equation, $\cos(e^x)$. The range of the cosine function is limited to $[-1, 1]$, which means the maximum value $\cos(e^x)$ can attain is 1.
From our previous finding that: $$ 2^x + 2^{-x} \geq 2 $$ we conclude that the expression $2^x + 2^{-x}$ is always greater than or equal to 2, which is already outside the maximum range of $\cos(e^x)$. Therefore, the two sides of the equation cannot be equal:
$$ \cos(e^x) \neq 2^x + 2^{-x} $$
Since this holds for any real $x$, there are no real solutions where $\cos(e^x) = 2^x + 2^{-x}$. Therefore, the number of solutions is:
(a) 0.
Which of the following are statements?
(A) $4 - x = 8$ (B) 4 is odd (C) $5 \in {1, 4, 5, 6}$ (D) $2 < 3$
To determine which of the options are statements in mathematics, recall that a statement is a claim that is either definitely true or definitely false, but not both. Let's review each option:
-
(A) $4 - x = 8$: This expression depends on the value of $x$. For example, it is true if $x = -4$ but false for any other value of $x$. Thus, because it is not universally true or false, it is not a statement.
-
(B) 4 is odd: This claim is always false, as 4 is an even number. Since this expression consistently evaluates to false, it is a statement.
-
(C) $5 \in {1, 4, 5, 6}$: This is saying that 5 is an element of the set ${1, 4, 5, 6}$. This is always true since 5 is indeed in the set. Hence, this is a statement.
-
(D) $2 < 3$: This claim is always true, making it a statement.
Summary: Among the given options, (B), (C), and (D) are statements as they consistently present a status (true or false), whereas (A) is not a statement since its truth varies with the value of $x$.
9 On the number line, the value of $(-3) \times 3$ lies on the right-hand side of: (a) -10 (b) -4 (c) 0 (d) 9
The expression given for evaluation is $$(-3) \times 3$$. The product of these numbers is: $$ -3 \times 3 = -9 $$ In analyzing where on the number line this value lies relative to the provided options, we find:
- The right-hand side of -10 is where -9 would be because on a number line, numbers increase in value from left to right. Hence, -9 is greater than -10 and would appear on the right side of it.
From the options:
- (a) -10
- (b) -4
- (c) 0
- (d) 9
Option (a) -10 is correct as -9 is indeed on the right-hand side of -10 on a number line.
For each positive real number $\lambda$, let $A_{\lambda}$ be the set of all natural numbers $n$ such that $|\sin (\sqrt{n-1}) - \sin (\sqrt{n})| < \lambda$. Let $A_{\lambda}^{c}$ be the complement of $A_{\lambda}$ in the set of all natural numbers. Then
A. $A_{\frac{1}{2}}, A_{\frac{1}{3}}, A_{\frac{2}{5}}$ are all finite sets.
B. $A_{\frac{1}{2}}$ is a finite set but $A_{\frac{1}{3}}, A_{\frac{2}{5}}$ are infinite sets.
C. $A_{\frac{1}{2}}^{c}, A_{\frac{1}{3}}^{c}, A_{\frac{2}{5}}^{c}$ are all finite sets.
D. $A_{\frac{1}{3}}, A_{\frac{2}{5}}$ are finite sets, but $A_{\frac{1}{2}}$ is an infinite set.
Solution
The correct option is C. $A_{\frac{1}{2}}^{c}, A_{\frac{1}{3}}^{c}, A_{\frac{2}{5}}^{c}$ are all finite sets.
As $n \rightarrow \infty$, we consider the expression: $$ |\sin (\sqrt{n-1}) - \sin (\sqrt{n})|. $$ By the properties of trigonometric functions and considering the Taylor expansion around large $n$, this absolute difference approaches 0. Thus, for large values of $n$, the expression: $$ |\sin (\sqrt{n-1}) - \sin (\sqrt{n})| $$ becomes very small and certainly less than any positive $\lambda$.
Consequently, there are infinite number of natural numbers $n$ satisfying: $$ |\sin (\sqrt{n-1}) - \sin (\sqrt{n})| < \lambda, $$ for any fixed positive $\lambda$. Therefore, the sets $A_{\frac{1}{2}}$, $A_{\frac{1}{3}}$, and $A_{\frac{2}{5}}$ are all infinite sets.
This implies that their complements, $A_{\frac{1}{2}}^{c}, A_{\frac{1}{3}}^{c}, A_{\frac{2}{5}}^{c}$, contain only those natural numbers that do not satisfy the condition, which are finite in number as the inequality holds for almost all large $n$. Thus, C is the correct option.
The equation $e^{\sin x} - e^{-\sin x} - 4=0$ has
A. an infinite number of real roots
B. no real roots
C. exactly one real root
D. exactly four real roots
The correct answer is Option B: no real roots.
To analyze the given equation, we start by simplifying $$ e^{\sin x} - e^{-\sin x} - 4 = 0. $$ This expression can be re-written in terms of the hyperbolic sine function as: $$ e^{\sin x} - e^{-\sin x} = 2\sinh(\sin x). $$ So, the equation simplifies to: $$ 2\sinh(\sin x) - 4 = 0. $$ However, let's solve it directly without re-phrasing in terms of hyperbolic functions: $$ e^{\sin x} - e^{-\sin x} = 4, $$ which means: $$ e^{2\sin x} - 4e^{\sin x} - 1 = 0. $$ Let $u = e^{\sin x}$. We then rewrite the equation as: $$ u^2 - 4u - 1 = 0. $$ Using the quadratic formula, we find the solutions for $u$: $$ u = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 \pm \sqrt{5}. $$
Analyzing Possible Values of $u$:
-
If $u = 2 + \sqrt{5}$:
$$ e^{\sin x} = 2 + \sqrt{5}. $$ Solving for $\sin x$: $$ \sin x = \ln(2 + \sqrt{5}). $$ Note that $\ln(2 + \sqrt{5})$ is positive and exceeds the maximum possible value of sine, which is $1$. Hence, $\sin x$ cannot take this value.
-
If $u = 2 - \sqrt{5}$:
$$ e^{\sin x} = 2 - \sqrt{5}. $$ Note that $2 - \sqrt{5}$ is less than $1$, and in fact, negative, which is outside the range of the exponential function $e^{\sin x}$ (which is always positive). Hence, the natural logarithm of a negative number is not defined in the real number system.
Thus, in both cases, $\sin x$ has no valid real values. Consequently, there are no real roots for the equation $e^{\sin x} - e^{-\sin x} - 4 = 0$.
Statement 'There are 23 hours in a day' is
A) always true
B) always false
C) ambiguous
D) cannot be proved
The correct answer is B) always false.
It is a well-established fact that a day consists of 24 hours. Therefore, the statement claiming there are 23 hours in a day is always false.
Which one of the following is not correct for the features of the exponential function given by $f(x)=b^{x}$, where $b>1$?
A. For very large negative values of $x$, the function is very close to 0.
B. The domain of the function is $\mathbb{R}$, the set of real numbers.
C. The point $(1,0)$ is always on the graph of the function.
D. The range of the function is the set of all positive real numbers.
The correct answer is C: The point $(1,0)$ is always on the graph of the function.
Explanation:
-
Option A: Correct. When $x$ takes on very large negative values, $b^{x}$ approaches zero since $b > 1$. For example, if $b = 2$ and $x = -10$, $2^{-10} = \frac{1}{2^{10}} = \frac{1}{1024}$, which is very close to 0.
-
Option B: Correct. The domain of the exponential function $f(x) = b^x$ is all real numbers, $\mathbb{R}$. This is because exponential functions are defined for any real exponent.
-
Option C: Incorrect. The point $(1,0)$ would mean that $f(1) = 0$. However, since $b > 1$, any positive exponent of $b$ will also be greater than 1 (not zero). Thus, the point on the graph for $x = 1$ is $(1, b)$, not $(1, 0)$. The usual fixed point for such functions is $(0, 1)$, where $f(0) = b^0 = 1$.
-
Option D: Correct. The range of the function, given $b^x$ where $b > 1$, is the set of all positive real numbers $(0, \infty)$. The exponential function never reaches zero or negative values and grows infinitely as $x$ increases.
Clearly, the incorrect feature here is mentioned in Option C.
If 0.3528 is expressed in the form of $\frac{p}{2^{m} \cdot 5^{n}}$, find the smallest values of $m$, $n$, and $p$.
To express $0.3528$ in the form $\frac{p}{2^m \cdot 5^n}$ and find the values of $p$, $m$, and $n$, we start by converting $0.3528$ into a fraction.
Step 1: Convert to a Fraction
We convert $0.3528$ to a fraction by counting the digits after the decimal point. Since there are four digits, $0.3528$ can be written as: $$ \frac{3528}{10000} $$
Step 2: Simplify the Fraction
Next, we need to simplify the fraction $\frac{3528}{10000}$. Both the numerator and the denominator can be simplified by dividing by their greatest common divisor.
The number $10000$ factors into $2^4 \cdot 5^4$.
We simplify $3528$ by finding its prime factors (using successive division by small prime numbers), but for this context, if we apply simplification directly: $$ \text{Common factors of } 3528 \text{ and } 10000 \text{ are } 2^3 \text{ and } 5^3. $$ So the simplified version: $$ \frac{441}{1250} $$
Step 3: Factorize the Denominator
We further express $1250 = 2 \cdot 5^4$. This gives us the structure needed for $\frac{p}{2^m \cdot 5^n}$.
Step 4: Identify Values of $p$, $m$, and $n$
From $\frac{441}{2 \cdot 5^4}$, we can now easily express it in the required format.
The smallest values are:
$p = 441$
$m = 1$ (from the factor of $2^1$)
$n = 4$ (from the factor of $5^4$)
Thus, $0.3528$ expressed as $\frac{p}{2^m \cdot 5^n}$ with the smallest values for $p$, $m$, and $n$ is $\frac{441}{2^1 \cdot 5^4}$.
A natural number when increased by 12, equals 160 times its reciprocal. Find the number.
From the problem, we are given that a natural number, when increased by 12, equals 160 times its reciprocal. Let's denote the natural number as $a$. Based on the given statement, we can set up the following equation:
$$ a + 12 = 160 \times \frac{1}{a} $$
To eliminate the fraction, we'll multiply both sides of the equation by $a$:
$$ a(a + 12) = 160 $$
Expanding the left side of the equation, we get:
$$ a^2 + 12a = 160 $$
We can then move all terms to one side of the equation to set the equation to zero:
$$ a^2 + 12a - 160 = 0 $$
Now, let's factorize this quadratic equation:
$$ a^2 + 20a - 8a - 160 = 0 $$ $$ (a + 20)(a - 8) = 0 $$
Setting each factor equal to zero gives us two potential solutions for $a$:
$$ a + 20 = 0 \quad \text{or} \quad a - 8 = 0 $$ $$ a = -20 \quad \text{or} \quad a = 8 $$
However, since $a$ is a natural number, it must be positive. Therefore, $a = -20$ is not a valid solution.
Thus, the natural number is:
$$ \textbf{a = 8} $$
A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.
Let's solve the given problem step by step.
Problem Statement:
We have a two-digit number $10x + y$, and we know:
The quotient when this number is divided by the sum of its digits $(x+y)$ is 6, with no remainder.
Subtracting 9 from this number results in the digits being reversed.
Breaking Down the Problem:
Step 1: Equation from division quotientGiven the quotient is 6, we can form the equation: $$ \frac{10x + y}{x + y} = 6 $$ Multiplying both sides by $(x+y)$, we get: $$ 10x + y = 6(x + y) $$ Solving for $x$ and $y$: $$ 10x + y = 6x + 6y $$ $$ 4x = 5y $$ This simplifies to: $$ x = \frac{5y}{4} $$
Step 2: Reversed digits after subtractionWhen subtracting 9 from the number, its digits are reversed. Thus, we can write: $$ 10x + y - 9 = 10y + x $$ $$ 10x + y - 9 = 10y + x $$ Rearranging terms gives: $$ 9x - 9y = 9 $$ $$ x - y = 1 $$
Step 3: Solving the system of equationsWe have the equations:
$x = \frac{5y}{4}$
$x - y = 1$
Substitute $x$ from the first equation into the second: $$ \frac{5y}{4} - y = 1 $$ $$ \frac{5y - 4y}{4} = 1 $$ $$ \frac{y}{4} = 1 $$ $$ y = 4 $$
Using the value of $y$ in $x - y = 1$: $$ x = y + 1 = 4 + 1 = 5 $$
Step 4: Formulate the original numberSince $x=5$ and $y=4$, the number is: $$ 10x + y = 10 \times 5 + 4 = 54 $$
Conclusion: The original number, which satisfies all given conditions, is 54.
Show that any number of the form $4^{n}$, where $n \in \mathbb{N}$, can never end with the digit 0.
To prove that any number of the form (4^n), where (n) is a natural number, can never end with the digit 0, we can analyze the units (last) digit of (4^n) for different values of (n).
Starting small: Examine the powers of 4:
(4^1 = 4): The units digit is 4.
(4^2 = 16): The units digit is 6.
(4^3 = 64): The units digit is 4.
(4^4 = 256): The units digit is 6.
Pattern Recognition: From the small cases above, observe the alternating pattern in the units digits:
If (n) is odd ((n=1, 3, 5, \dots)), (4^n) ends in 4.
If (n) is even ((n=2, 4, 6, \dots)), (4^n) ends in 6.
Understanding why the ending digit can't be 0:
Any number that ends with 0 must be divisible by 10. However, from the powers computed, (4^n) alternates between ending in 4 and 6, neither of which are divisible by 10.
This pattern holds true for all natural numbers since the units digit cycles through 4 and 6 due to the base number being 4 and 6 (from (4^1) and (4^2)).
Given this recurring pattern, we can conclude that neither 4 nor 6 is divisible by 10, confirming that (4^n) will never end with the digit 0 for any natural number (n).
Given that $\sqrt{5}$ is irrational, we can assume that $2 \sqrt{5}$ is also irrational because irrational numbers are closed under multiplication with rationals. Therefore, if we subtract a rational number such as 3 from an irrational number like $2 \sqrt{5}$, the result $2 \sqrt{5} - 3$ must be irrational as well. Thus, we have proved that $2 \sqrt{5}-3$ is an irrational number.
To demonstrate that $2 \sqrt{5} - 3$ is irrational, we start by acknowledging the given that $\sqrt{5}$ is irrational. The problem asks us to establish the irrationality of $2 \sqrt{5} - 3$.
Assumption: Make an assumption that contradicts what we want to prove for the sake of argument. Suppose $2 \sqrt{5} - 3$ is a rational number.
Rational Number Definition: Recall that a rational number can be expressed as $\frac{p}{q}$, where $p$ and $q$ are integers with no common factors other than 1, and $q \neq 0$.
Manipulation:
If $2 \sqrt{5} - 3$ is rational, we could write it as $\frac{p}{q}$,
Then, rearranging gives us $2\sqrt{5} = \frac{p}{q} + 3$,
Simplifying further, $\sqrt{5} = \frac{p + 3q}{2q}$.
Contradiction Arises: Now notice, $\sqrt{5}$ is on one side of the equation presented as a quotient of integers which implies it's rational, but we were given that $\sqrt{5}$ is irrational. This discrepancy leads to a contradiction.
Conclusion:
Since our assumption (that $2 \sqrt{5} - 3$ is rational) leads to a contradiction, it must be false. Hence, $2\sqrt{5} - 3$ cannot be rational, meaning it is irrational.
Therefore, by assuming $2 \sqrt{5} - 3$ is rational and arriving at a contradiction, we have proved that the expression indeed must be irrational. This technique, known as a proof by contradiction, robustly establishes that $2 \sqrt{5} - 3$ is irrational.
Find after how many places of the decimal the decimal form of the number $\frac{27}{2^{3} \cdot 5^{4} \cdot 3^{2}}$ will terminate.
To determine how many places after the decimal the decimal form of the fraction $$\frac{27}{2^3 \cdot 5^4 \cdot 3^2}$$ will terminate, we first simplify the fraction by canceling out any common factors in the numerator and the denominator.
Start by examining the exponent of each prime factor in the fraction:
The numerator, 27, can be expressed as $3^3$.
The denominator is $2^3 \cdot 5^4 \cdot 3^2$.
So, the fraction becomes: $$ \frac{3^3}{2^3 \cdot 5^4 \cdot 3^2} = \frac{3^3}{3^2 \cdot 2^3 \cdot 5^4} = \frac{3^{3-2}}{2^3 \cdot 5^4} = \frac{3^1}{2^3 \cdot 5^4} = \frac{3}{2^3 \cdot 5^4}. $$ Here, we reduced $3^3$ and $3^2$ to $3^1$ by subtracting their exponents.
Since the fraction now only involves prime factors 2 and 5 in the denominator (post simplification), it will result in a terminating decimal. A decimal terminates if and only if the denominator, after reducing any common factors with the numerator, can be expressed as a product of only prime factors 2 and/or 5.
The decimal representation of this faction will terminate after calculating the decimal position as equivalent to the higher power of 2 or 5 in the denominator:
Here, the highest power in the denominator is the power of 5, which is 4.
Therefore, the decimal form of $$\frac{27}{2^3 \cdot 5^4 \cdot 3^2}$$ will terminate after 4 decimal places.
Prove that $\sqrt{3}$ is an irrational number.
To prove that $\sqrt{3}$ is irrational, we use the method of contradiction. Let's start by assuming the opposite, that $\sqrt{3}$ is rational. If $\sqrt{3}$ is rational, this means it can be expressed as a ratio of two integers, say $\frac{p}{q}$, where $p$ and $q$ are integers with no common factors (i.e., they are coprime), and $q$ is not zero:
$$ \sqrt{3} = \frac{p}{q} $$
Squaring both sides, we get:
$$ 3 = \frac{p^2}{q^2} $$
Rearranging gives:
$$ 3q^2 = p^2 $$
This equation tells us that $p^2$ is divisible by $3$. Because $3$ is a prime number, it follows from the fundamental theorem of arithmetic that $p$ must also be divisible by $3$. Let’s say $p = 3k$, where $k$ is an integer.
Substituting $p = 3k$ into the equation $3q^2 = p^2$, we get:
$$ 3q^2 = (3k)^2 = 9k^2 $$
Dividing both sides by $3$, we then have:
$$ q^2 = 3k^2 $$
Now, this new equation tells us that $q^2$ is divisible by $3$, and again, by the fundamental theorem of arithmetic, $q$ must also be divisible by $3$. However, we initially assumed that $p$ and $q$ have no common factors (meaning they are coprime). The fact that both $p$ and $q$ are divisible by $3$ contradicts this assumption.
The contradiction arises from our initial, incorrect assumption that $\sqrt{3}$ is rational. Therefore, we are forced to conclude that $\sqrt{3}$ is irrational. This completes the proof.
After how many places will the decimal form of the number $\frac{27}{2^{3} 5^{4} 3^{2}}$ terminate?
A) 1
B) 2
C) 3
D) 4
To determine after how many places the decimal form of the number $$ \frac{27}{2^{3} 5^{4} 3^{2}} $$ will terminate, we first need to understand the concept of decimal termination.
A fraction in its lowest terms will have a terminating decimal if and only if the denominator is of the form $2^m5^n$, where $m$ and $n$ are non-negative integers. This means all the prime factors of the denominator should only be 2 or 5.
Let's simplify the given expression: $$ \frac{27}{2^3 5^4 3^2}. $$
Here:
The numerator is 27.
The denominator is $2^3 \cdot 5^4 \cdot 3^2$.
Note that the presence of $3^2$ in the denominator suggests that the fraction is not in its simplest form to easily determine if the decimal terminates. First, we factor out 27 from the numerator and the $3^2$ in the denominator, which simplifies to 9: $$\frac{27}{9} = 3.$$
So, the simplified expression of the fraction is: $$ \frac{3}{2^3 5^4}. $$
In this case:
$2^3$ contributes 3 to the denominator.
$5^4$ contributes 4 to the denominator.
According to the rule stated earlier, the higher power of 2 or 5 in the denominator will determine the number of decimal places. Here, the highest power is 4 (from $5^4$). Therefore, the decimal form of the number will terminate after 4 places.
So, the correct answer is D) 4. The decimal terminates after 4 places since that is the higher power among the distractors in the denominator after simplification.
For a rational number $\frac{p}{q}$ to be a terminating decimal, the denominator $q$ must be of the form $2^{m} \times 5^{n}$, where $m, n$ are
A. Integers B. Natural numbers C. Positive integers D. Non-negative integers.
In order to determine if a rational number $\frac{p}{q}$ is a terminating decimal, we need to consider the denominator $q$ in its prime factorized form. Specifically, $q$ must be expressible as $2^m \times 5^n$, where $m$ and $n$ can take specific values depending on the conditions for termination.
Key Concept:
A decimal representation of a fraction $\frac{p}{q}$ terminates if and only if the prime factorization of the denominator $q$ includes only the prime numbers 2 and 5. This is because the base of our number system (decimal) is 10, which is $2 \times 5$. Other factors in the denominator would result in repeating decimals.
Detailing $m$ and $n$:
When we express $q$ as $2^m \times 5^n$:
$m$ is the exponent corresponding to the factor 2.
$n$ is the exponent corresponding to the factor 5.
Both $m$ and $n$ need to be such that they can accurately represent the prime factors in $q$. This leads us to the types of number $m$ and $n$ can be:
Natural numbers include all positive integers (1, 2, 3, ...). These are sufficient to count occurrences of a factor in a number.
Zero is also important as $2^0 \times 5^0 = 1$, which represents a scenario where either factor might not actually be present in the denominator but is theoretically counted as present zero times. Thus, including zero is necessary to cover all possible values of $q$ that would result in a terminating decimal.
Conclusion:
Given our need to include zero, the best categorization for $m$ and $n$ is as non-negative integers. Non-negative integers include all the natural numbers and zero, thus perfectly fitting the requirement for $m$ and $n$.
Therefore, for the rational number $\frac{p}{q}$ to be a terminating decimal, the denominator $q$ must be of the form $2^m \times 5^n$, where $m$ and $n$ are:
D. Non-negative integers.
If $\frac{1}{\log_{3} \pi} + \frac{1}{\log_{4} \pi} > x$, then $x$ is equal to:
A) 2
B) 3
C) 4
D) 5
Given the inequality: $$ \frac{1}{\log_{3} \pi} + \frac{1}{\log_{4} \pi} > x $$
We can rewrite the inequality by using the change of base formula. Let's transform the logarithms: $$ \frac{1}{\log_{3} \pi} = \log_{\pi} 3 \quad \text{and} \quad \frac{1}{\log_{4} \pi} = \log_{\pi} 4 $$
Therefore, the inequality becomes: $$ \log_{\pi} 3 + \log_{\pi} 4 > x $$
By using the property of logarithms which states that $\log_b m + \log_b n = \log_b (mn)$, we can combine the terms on the left: $$ \log_{\pi} 3 + \log_{\pi} 4 = \log_{\pi} (3 \times 4) = \log_{\pi} 12 $$
Thus, the inequality now is: $$ \log_{\pi} 12 > x $$
Next, we need to determine the value of $\log_{\pi} 12$. We know that $\pi \approx 3.14$, so we have: $$ \pi^2 < 12 < \pi^3 $$
Taking the logarithm base $\pi$ for these inequalities: $$ \log_{\pi} (\pi^2) < \log_{\pi} 12 < \log_{\pi} (\pi^3) $$
Using the logarithm properties again, we get: $$ 2 < \log_{\pi} 12 < 3 $$
Therefore: $$ \log_{\pi} 12 > 2 $$
So, the value of $x$ must be: $$ x = 2 $$
Hence, the correct answer is:
A) 2
💡 Have more questions?
Ask Chatterbot AI