Squares and Square Roots

Square Numbers or Perfect Squares: Definition and Properties

Defining Square Numbers

In mathematics, the operation of squaring a number means multiplying the number by itself. When an integer is squared, the result is a special type of number called a square number or a perfect square.

Formally, a positive integer $n$ is a perfect square if it can be expressed as the square of another integer $m$. That is, $n = m \times m$, or in exponential notation, $n = m^2$, for some integer $m$. While $m$ can be any integer (positive, negative, or zero), the perfect square $n$ is typically understood to be the non-negative result obtained by squaring a whole number ($m \in \{0, 1, 2, 3, \ldots\}$).

Examples of perfect squares obtained by squaring the first few whole numbers:

• $0^2 = 0 \times 0 = 0$. So, 0 is a perfect square.
• $1^2 = 1 \times 1 = 1$. So, 1 is a perfect square.
• $2^2 = 2 \times 2 = 4$. So, 4 is a perfect square.
• $3^2 = 3 \times 3 = 9$. So, 9 is a perfect square.
• $4^2 = 4 \times 4 = 16$. So, 16 is a perfect square.
• $5^2 = 5 \times 5 = 25$. So, 25 is a perfect square.
• $10^2 = 10 \times 10 = 100$. So, 100 is a perfect square.
• $15^2 = 15 \times 15 = 225$. So, 225 is a perfect square.

The sequence of the first few non-negative perfect squares is $\mathbf{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, \ldots}$.

Squaring a negative integer also results in a positive perfect square. For example, $(-4)^2 = (-4) \times (-4) = 16$. This is why every positive perfect square greater than 0 has two integer square roots: one positive and one negative (e.g., the square roots of 16 are 4 and -4). However, when we refer to "a perfect square", we are usually talking about the non-negative result.

Properties Characterizing Perfect Squares

Perfect squares have several distinct properties related to their digits and their prime factorizations that can help in identifying whether a number is a perfect square without calculating its square root:

1. Ending Digits (Unit's Place):

   The last digit (the digit in the units place) of a perfect square can only be one of the digits $\mathbf{0, 1, 4, 5, 6,}$ or $\mathbf{9}$. This means a number ending in $2, 3, 7,$ or $8$ can never be a perfect square.

   Explanation: The last digit of a product of two integers is determined solely by the product of their last digits. So, the last digit of $m^2$ is the same as the last digit of (last digit of $m$)$^2$. We only need to check the squares of the possible last digits of any integer (0 through 9):

   $$ 0^2 = 0 \quad (\text{ends in } 0) $$ $$ 1^2 = 1 \quad (\text{ends in } 1) $$ $$ 2^2 = 4 \quad (\text{ends in } 4) $$ $$ 3^2 = 9 \quad (\text{ends in } 9) $$ $$ 4^2 = 16 \quad (\text{ends in } 6) $$ $$ 5^2 = 25 \quad (\text{ends in } 5) $$ $$ 6^2 = 36 \quad (\text{ends in } 6) $$ $$ 7^2 = 49 \quad (\text{ends in } 9) $$ $$ 8^2 = 64 \quad (\text{ends in } 4) $$ $$ 9^2 = 81 \quad (\text{ends in } 1) $$

   The possible last digits of a perfect square are thus $\{0, 1, 4, 5, 6, 9\}$. If a number ends in any other digit, it cannot be a perfect square.

   Example: $123$ ends in $3$, so it's not a perfect square. $457$ ends in $7$, not a perfect square. $1028$ ends in $8$, not a perfect square. $5002$ ends in $2$, not a perfect square.

   Note: This is a necessary condition, but not sufficient. A number ending in 0, 1, 4, 5, 6, or 9 is not guaranteed to be a perfect square (e.g., 10 ends in 0, but it's not a perfect square).

2. Number of Zeros at the End:

   If a perfect square ends with one or more zeros, the number of trailing zeros must always be an even number (2, 4, 6, ...).

   Example: $100$ ends in 2 zeros ($10^2$). $400$ ends in 2 zeros ($20^2$). $90000$ ends in 4 zeros ($300^2$). $1,000$ ends in 3 zeros, so it cannot be a perfect square. $10,000$ ends in 4 zeros ($100^2$).

   Explanation: A trailing zero in a number comes from a factor of 10. $10 = 2 \times 5$. If a number $m$ ends in $k$ zeros, its prime factorization contains $2^k$ and $5^k$ as factors (or higher powers of 2 or 5, but at least $k$ factors of each). When we square $m$, $m^2 = (p_1^{e_1} \ldots 2^k 5^k \ldots)^2 = p_1^{2e_1} \ldots 2^{2k} 5^{2k} \ldots$. The minimum power of 2 and 5 in the factorization of $m^2$ (ignoring primes other than 2 and 5) will be $2k$ (if $m$ itself didn't have extra factors of 2 or 5 beyond those forming the trailing zeros). The number of trailing zeros is determined by the minimum of the powers of 2 and 5 in its prime factorization. For a perfect square, these minimum powers must be even, resulting in an even number of trailing zeros.

3. Sum of Consecutive Odd Numbers:

   A number is a perfect square if and only if it can be expressed as the sum of a consecutive sequence of positive odd numbers starting from $1$. The sum of the first $n$ consecutive positive odd numbers is equal to $n^2$.

   $$ 1 = 1^2 $$ $$ 1 + 3 = 4 = 2^2 $$ $$ 1 + 3 + 5 = 9 = 3^2 $$ $$ 1 + 3 + 5 + 7 = 16 = 4^2 $$ $$ 1 + 3 + 5 + \ldots + (2n-1) = n^2 $$

   This property can be used to check if a number is a perfect square by repeatedly subtracting consecutive odd numbers (starting from 1) and checking if the result is 0. This forms the basis of the repeated subtraction method for finding square roots.

4. Prime Factorization:

   In the unique prime factorization of a perfect square, the exponent of every prime factor is always an even number.

   Explanation: If a positive integer $n$ is a perfect square, then $n = m^2$ for some integer $m$. Let the prime factorization of $m$ be $m = p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k}$, where $p_i$ are distinct primes and $e_i \ge 0$ are integers. Then the prime factorization of $n$ is obtained by squaring the factorization of $m$: $n = (p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k})^2 = p_1^{2e_1} p_2^{2e_2} \ldots p_k^{2e_k}$. Each exponent in the factorization of $n$ is of the form $2e_i$, which is an even integer.

   Conversely, if the prime factorization of a number $n$ has all even exponents, $n = p_1^{2e_1} p_2^{2e_2} \ldots p_k^{2e_k}$. We can rewrite this as $n = (p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k})^2$. Since $e_i$ are integers, $m = p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k}$ is also an integer, and $n = m^2$ is a perfect square.

   Example: $64 = 2^6$. The exponent is 6 (even). $64 = (2^3)^2 = 8^2$.

   Example: $225 = 3^2 \times 5^2$. The exponents are 2 and 2 (both even). $225 = (3^1 \times 5^1)^2 = 15^2$.

   Example: $784$. Prime factorization of 784: $784 = 2^4 \times 7^2$. Exponents are 4 and 2 (both even). $784 = (2^2 \times 7^1)^2 = (4 \times 7)^2 = 28^2$.

   Example: $12 = 2^2 \times 3^1$. The exponent of 3 is 1 (odd). So 12 is not a perfect square.

5. Between Consecutive Squares:

   Between the squares of two consecutive integers, $n^2$ and $(n+1)^2$, there are exactly $2n$ non-perfect square integers.

   Explanation: The integers between $n^2$ and $(n+1)^2$ (exclusive of the endpoints) are $n^2+1, n^2+2, \ldots, (n+1)^2-1$. The number of integers in this range is $((n+1)^2 - 1) - (n^2 + 1) + 1$.
   $(n+1)^2 - 1 = (n^2 + 2n + 1) - 1 = n^2 + 2n$.
   The integers are from $n^2+1$ to $n^2+2n$. The count is $(n^2+2n) - (n^2+1) + 1 = n^2 + 2n - n^2 - 1 + 1 = 2n$.
   None of these $2n$ integers can be perfect squares, because if $k^2$ was one of them, then $n^2 < k^2 < (n+1)^2$. Taking the square root would give $n < k < n+1$. But there is no integer $k$ strictly between two consecutive integers $n$ and $n+1$. Thus, the integers between $n^2$ and $(n+1)^2$ are all non-perfect squares.

   Example: Between $5^2 = 25$ and $6^2 = 36$. The number of non-perfect squares is $2 \times 5 = 10$. These are $26, 27, 28, 29, 30, 31, 32, 33, 34, 35$.

These properties are valuable for quickly identifying perfect squares and form the basis for methods used to find square roots.

Pythagorean Triplets

Definition based on the Pythagorean Theorem

A Pythagorean triplet (or Pythagorean triple) is a set of three positive integers, commonly denoted as $(a, b, c)$, that satisfy the equation $a^2 + b^2 = c^2$. This definition is a direct consequence of the Pythagorean theorem, which describes the relationship between the side lengths of a right-angled triangle.

The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle, which is the longest side) is equal to the sum of the squares of the lengths of the other two sides (called the legs or cathetus). If the lengths of the legs are $a$ and $b$, and the length of the hypotenuse is $c$, the theorem is expressed by the equation:

A Pythagorean triplet $(a, b, c)$ therefore represents three positive integers that can be the side lengths of a right-angled triangle. In such a triplet, $c$ is always the largest number, representing the length of the hypotenuse, while $a$ and $b$ represent the lengths of the legs.

Examples of Pythagorean Triplets

The most famous and smallest set of positive integers that form a Pythagorean triplet is $(3, 4, 5)$. Let's verify this using the Pythagorean theorem:

Let $a=3, b=4, c=5$.

Since $a^2 + b^2 = c^2$ ($25 = 25$), the set of integers $(3, 4, 5)$ is a Pythagorean triplet. This means a right-angled triangle can have side lengths of 3 units, 4 units, and 5 units.

Other examples of Pythagorean triplets (check them by squaring the first two numbers and summing, then squaring the third number):

• $(5, 12, 13)$: $5^2 + 12^2 = 25 + 144 = 169$. $13^2 = 169$. $169 = 169$.
• $(8, 15, 17)$: $8^2 + 15^2 = 64 + 225 = 289$. $17^2 = 289$. $289 = 289$.
• $(7, 24, 25)$: $7^2 + 24^2 = 49 + 576 = 625$. $25^2 = 625$. $625 = 625$.
• $(20, 21, 29)$: $20^2 + 21^2 = 400 + 441 = 841$. $29^2 = 841$. $841 = 841$.

A useful property is that any positive integer multiple of a Pythagorean triplet is also a Pythagorean triplet. If $(a, b, c)$ is a Pythagorean triplet, then for any positive integer $k$, the set $(ka, kb, kc)$ is also a Pythagorean triplet.

Proof: Given $a^2 + b^2 = c^2$. Consider the set $(ka, kb, kc)$. We check if $(ka)^2 + (kb)^2 = (kc)^2$.

Factor out $k^2$:

Since $(a, b, c)$ is a triplet, we know $a^2 + b^2 = c^2$. Substitute this into the equation:

Rewrite the right side as a square:

So, $(ka)^2 + (kb)^2 = (kc)^2$, which means $(ka, kb, kc)$ is a Pythagorean triplet.

Example: Take the triplet $(3, 4, 5)$ and $k=2$. The new set is $(2 \times 3, 2 \times 4, 2 \times 5) = (6, 8, 10)$. Check: $6^2 + 8^2 = 36 + 64 = 100$. $10^2 = 100$. $100 = 100$, so $(6, 8, 10)$ is a Pythagorean triplet.

Primitive Pythagorean Triplets

A Pythagorean triplet $(a, b, c)$ is called a primitive Pythagorean triplet if the three integers $a, b,$ and $c$ have no common factors other than $1$. In other words, their Greatest Common Divisor (GCD) is $1$, i.e., $\text{GCD}(a, b, c) = 1$. Primitive triplets are the basic, non-reducible sets of integer side lengths for right triangles. Any non-primitive triplet is a positive integer multiple of a primitive triplet.

Examples of primitive triplets: $(3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25)$. For each of these, the GCD of the three numbers is 1.

Examples of non-primitive triplets: $(6, 8, 10)$. $\text{GCD}(6, 8, 10) = 2$. This is the primitive triplet $(3, 4, 5)$ multiplied by 2. $(9, 12, 15)$. $\text{GCD}(9, 12, 15) = 3$. This is the primitive triplet $(3, 4, 5)$ multiplied by 3.

Generating Pythagorean Triplets

There are formulas to generate Pythagorean triplets. A well-known formula, attributed to Euclid, can generate all primitive Pythagorean triplets under specific conditions and all non-primitive triplets under slightly relaxed conditions.

Euclid's Formula for Generating Triplets: Given two positive integers $m$ and $n$ with $m > n$, the integers $a, b,$ and $c$ calculated as follows form a Pythagorean triplet $(a, b, c)$:

To generate primitive Pythagorean triplets using this formula, $m$ and $n$ must satisfy the additional conditions:

• $m$ and $n$ must be co-prime (their Greatest Common Divisor must be 1, $\text{GCD}(m, n) = 1$).
• One of $m$ and $n$ must be an even integer, and the other must be an odd integer. (If both were odd, $m^2, n^2, m^2+n^2, m^2-n^2$ would be even, and $2mn$ would be even, so the triplet would be non-primitive, divisible by 2).

Verification of the formula: We need to check if $(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$.

Left side:

Combine like terms:

Right side:

Since the left side equals the right side ($m^4 + 2m^2 n^2 + n^4$), the integers $(m^2 - n^2, 2mn, m^2 + n^2)$ satisfy the Pythagorean theorem and thus form a Pythagorean triplet for any integers $m > n > 0$.

Examples of generating primitive triplets using $m$ and $n$ satisfying the conditions ($m > n$, $\text{GCD}(m,n)=1$, one even, one odd):

• Let $m=2, n=1$. Conditions: $2>1$, $\text{GCD}(2,1)=1$, 2 is even, 1 is odd. All satisfied.
  $a = 2^2 - 1^2 = 4 - 1 = 3$
  $b = 2(2)(1) = 4$
  $c = 2^2 + 1^2 = 4 + 1 = 5$
  Triplet: $\mathbf{(3, 4, 5)}$.
• Let $m=4, n=1$. Conditions: $4>1$, $\text{GCD}(4,1)=1$, 4 is even, 1 is odd. All satisfied.
  $a = 4^2 - 1^2 = 16 - 1 = 15$
  $b = 2(4)(1) = 8$
  $c = 4^2 + 1^2 = 16 + 1 = 17$
  Triplet: $(15, 8, 17)$. Usually written ordered by size: $\mathbf{(8, 15, 17)}$.
• Let $m=4, n=3$. Conditions: $4>3$, $\text{GCD}(4,3)=1$, 4 is even, 3 is odd. All satisfied.
  $a = 4^2 - 3^2 = 16 - 9 = 7$
  $b = 2(4)(3) = 24$
  $c = 4^2 + 3^2 = 16 + 9 = 25$
  Triplet: $\mathbf{(7, 24, 25)}$.

If $m$ and $n$ do not satisfy the primitive conditions, the formula will produce a non-primitive triplet. For example, if $m=3, n=1$ (both odd, GCD is 1, but not one even/one odd): $a = 3^2 - 1^2 = 8$, $b = 2(3)(1) = 6$, $c = 3^2 + 1^2 = 10$. Triplet $(8, 6, 10)$, which is $2 \times (4, 3, 5)$, a non-primitive triplet based on $(3, 4, 5)$. If $m=6, n=3$ (GCD is 3, not 1): $a = 6^2 - 3^2 = 36 - 9 = 27$, $b = 2(6)(3) = 36$, $c = 6^2 + 3^2 = 36 + 9 = 45$. Triplet $(27, 36, 45)$, which is $9 \times (3, 4, 5)$, a non-primitive triplet.

Pythagorean triplets are fundamental in geometry, especially in construction, and are a topic of study in number theory (Diophantine equations).

Square Roots: Definition and Methods (Repeated Subtraction, Prime Factorisation, Division)

Understanding Square Roots

The operation of finding the square root is the inverse operation of squaring a number. Squaring a number $m$ means multiplying it by itself to get $x$ ($m^2 = x$). Finding the square root of $x$ means finding the number $m$ that was squared to produce $x$.

The square root of a number $x$ is any number $m$ such that $m^2 = x$.

• For example, the square roots of 25 are 5 (since $5^2 = 25$) and -5 (since $(-5)^2 = 25$).
• The square roots of 49 are 7 and -7.

Every positive real number $x$ has exactly two real square roots: one positive and one negative. The number 0 has only one square root, which is 0 itself ($\sqrt{0}=0$).

The symbol $\sqrt{\phantom{x}}$, called the radical symbol or radical sign, is used to denote the principal square root. The principal square root is the unique non-negative square root of a non-negative number.

• $\sqrt{25} = 5$ (This denotes only the positive square root)
• $-\sqrt{25} = -5$ (This denotes the negative square root)
• $\pm\sqrt{25} = \pm 5$ (This denotes both the positive and negative square roots)

The expression under the radical sign is called the radicand.

The square root of a negative number is not a real number because the square of any real number (whether positive, negative, or zero) is always non-negative. For example, $\sqrt{-9}$ is not a real number; it belongs to the set of complex numbers (specifically, it's a purely imaginary number, $3i$, where $i = \sqrt{-1}$).

If a positive integer is a perfect square (e.g., 4, 9, 16), its principal square root is a positive integer (e.g., $\sqrt{4}=2, \sqrt{9}=3, \sqrt{16}=4$). If a positive number is not a perfect square, its principal square root is a positive irrational number (e.g., $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{10}$).

Methods for Finding Square Roots

Various methods exist for finding the square root of a number. Some are suited specifically for perfect squares, while others are general methods for any non-negative number.

Method 1: Repeated Subtraction Method (Suitable for identifying small perfect squares)

This method is based on the property that the sum of the first $n$ consecutive positive odd numbers is equal to $n^2$. To find the square root of a given number, you repeatedly subtract consecutive positive odd numbers starting from 1. If the final result is 0, the original number is a perfect square, and the number of subtractions performed is its square root.

Steps:

1. Start with the given number.
2. Subtract the first positive odd number (1).
3. From the result, subtract the second positive odd number (3).
4. Continue subtracting consecutive positive odd numbers (5, 7, 9, ...) from the previous result.
5. If the result at any step is 0, the original number is a perfect square, and the number of steps (subtractions) performed is its principal square root. If you get a negative result at any step, the original number is not a perfect square.

Answer:

This method is good for understanding the concept but is inefficient for large numbers or checking if a number is not a perfect square.

Method 2: Prime Factorisation Method (Suitable for perfect squares)

This method utilizes the property that in the unique prime factorization of a perfect square, the exponent of every prime factor is always an even number. Finding the square root involves dividing each exponent by 2.

Steps:

1. Find the prime factorization of the given number. Express the factorization in exponential form.
2. Examine the exponents of all the prime factors. If all the exponents are even, the number is a perfect square. If even one exponent is odd, the number is not a perfect square.
3. To find the principal square root, divide each of the exponents in the prime factorization by $2$.
4. Multiply the prime factors raised to these new exponents. The result is the principal square root.

Answer:

Answer:

The prime factorization method is excellent for determining if a number is a perfect square and for simplifying square roots of perfect squares or numbers that are products of perfect squares and other factors.

Method 3: Long Division Method (General Method for any Non-Negative Number)

The long division method for finding square roots is an algorithm that can find the exact square root of a perfect square or provide a decimal approximation for the square root of any non-negative number to any desired level of precision. It is a systematic way to find the digits of the square root one by one.

Steps to find the square root of a number $N$:

1. Pair the Digits: Starting from the decimal point, group the digits of the number into pairs. For the integer part, group digits from right to left. For the decimal part, group digits from left to right. If the leftmost group in the integer part is a single digit, treat it as a pair. Add trailing zeros in pairs in the decimal part if you need to calculate decimal places in the square root. Place a bar (vinculum) over each pair of digits (and the single leftmost digit if applicable).
2. Find the First Digit: Consider the leftmost pair (or single digit). Find the largest digit whose square is less than or equal to this first group. This digit is the first digit of the square root. Write this digit as the first digit of the quotient (the square root) and also as the divisor for the first step. Subtract the square of this digit from the first group.
3. Bring Down and Double: Bring down the next pair of digits to the right of the current remainder. This forms the new dividend. Double the current quotient (the digits found so far) and write it to the left of the new dividend, leaving a blank space to its right. This is the start of the next trial divisor (if the current quotient is $q_c$, the trial divisor is $2q_c \_$).
4. Find the Next Digit: Find the largest digit ($d$) that can be placed in the blank space of the trial divisor such that when the resulting number ($2q_c \times 10 + d$) is multiplied by $d$, the product is less than or equal to the new dividend. This digit $d$ is the next digit of the square root. Write $d$ next to the current quotient and also in the blank space of the trial divisor. Subtract the product $(2q_c \times 10 + d) \times d$ from the new dividend.
5. Repeat: Bring down the next pair of digits and repeat steps 3 and 4. If you bring down a pair of digits from the decimal part of the original number, place a decimal point in the quotient at that step. Continue until you get a remainder of 0 (for perfect squares) or reach the desired number of decimal places.

Answer:

Answer:

The long division method is a systematic way to find the square root of any non-negative number to any desired precision, regardless of whether it is a perfect square or not.

Estimating Square Roots

Estimating the square root of a number is the process of finding an approximate value for its principal square root, especially when the number is not a perfect square. This skill is useful for quickly understanding the magnitude of a square root without performing a precise calculation, and it relies on knowing the values of common perfect squares and understanding the ordering of numbers on the number line.

Estimation using Consecutive Perfect Squares

The most fundamental way to estimate the square root of a non-negative number is to identify the two consecutive perfect squares between which the given number lies. The square root of the given number will then be located between the square roots of those two consecutive perfect squares.

Steps for Estimation:

1. Given a non-negative number $x$, find the two consecutive perfect squares, $n^2$ and $(n+1)^2$, such that $n^2 \le x < (n+1)^2$.
2. Take the square root of all parts of the inequality: $\sqrt{n^2} \le \sqrt{x} < \sqrt{(n+1)^2}$.
3. Simplify the perfect squares: $n \le \sqrt{x} < n+1$. This tells you that the integer part of $\sqrt{x}$ is $n$, and $\sqrt{x}$ lies between $n$ and $n+1$.
4. To get a rough sense of where $\sqrt{x}$ is between $n$ and $n+1$, consider whether $x$ is closer to $n^2$ or $(n+1)^2$. If $x$ is closer to $n^2$, $\sqrt{x}$ will be closer to $n$. If $x$ is closer to $(n+1)^2$, $\sqrt{x}$ will be closer to $n+1$.

Answer:

Answer:

Refining Estimates (using squares of decimals)

For a more refined estimate, you can extend the process by testing the squares of decimal numbers within the identified integer interval. For example, to refine the estimate for $\sqrt{7}$ (which is between 2 and 3):

• Try $2.5^2 = 6.25$. Since $6.25 < 7$, we know $\sqrt{7} > 2.5$.
• Try $2.6^2 = 6.76$. Since $6.76 < 7$, we know $\sqrt{7} > 2.6$.
• Try $2.7^2 = 7.29$. Since $7.29 > 7$, we know $\sqrt{7} < 2.7$.

So, we have narrowed down the estimate: $2.6 < \sqrt{7} < 2.7$. We can continue this process to get more decimal places:

• Try $2.64^2 = 6.9696$. Since $6.9696 < 7$, $\sqrt{7} > 2.64$.
• Try $2.65^2 = 7.0225$. Since $7.0225 > 7$, $\sqrt{7} < 2.65$.

So, $2.64 < \sqrt{7} < 2.65$. This provides an estimate of $\sqrt{7}$ as approximately 2.64 (or 2.65, depending on desired rounding). This process is essentially the manual digit-by-digit calculation that happens in the long division method for square roots or numerical approximation algorithms.

Estimating square roots provides a quick way to gauge the size of irrational square roots and is useful in various problem-solving contexts where an exact value is not required.

Square Roots of Decimals and Fractions

Finding the square roots of decimal numbers and fractions extends the concept of finding square roots of integers. The methods rely on the definition of the square root and properties of operations with decimals and fractions.

Square Roots of Decimal Numbers

When we square a decimal number, the number of decimal places in the result is double the number of decimal places in the original number. For example, $0.2 \times 0.2 = 0.04$ (1 decimal place becomes 2), $1.5 \times 1.5 = 2.25$ (1 becomes 2), $0.12 \times 0.12 = 0.0144$ (2 become 4). Conversely, when finding the square root of a perfect square decimal, the number of decimal places in the square root is half the number of decimal places in the original decimal. This implies that a perfect square decimal must have an even number of decimal places.

To find the principal square root of a perfect square decimal:

1. Count the number of decimal places in the given number. Let this be $m$. Verify that $m$ is an even number. If $m$ is odd, the number is not a perfect square of a number with a terminating decimal representation.
2. Remove the decimal point from the number. Find the principal square root of the resulting whole number using any suitable method (prime factorization, long division, or by inspection if it's a known square).
3. In the result obtained in step 2, insert the decimal point such that it has exactly $m/2$ decimal places, counting from the rightmost digit.

Answer:

Answer:

If the decimal number is not a perfect square, or if you are required to use a specific method like long division, that method can be applied directly to the decimal number. Remember to pair the decimal digits correctly (moving to the right from the decimal point) and place the decimal point in the quotient when you start dividing the decimal pairs.

Square Roots of Fractions

To find the principal square root of a fraction $\frac{a}{b}$ (where $a$ is a non-negative number and $b$ is a positive number), we can use the property that the square root of a quotient is equal to the quotient of the square roots. This property applies if the individual square roots are well-defined real numbers.

Steps:

1. Find the principal square root of the numerator, $\sqrt{a}$.
2. Find the principal square root of the denominator, $\sqrt{b}$.
3. Divide the result from step 1 by the result from step 2: $\frac{\sqrt{a}}{\sqrt{b}}$.
4. Simplify the resulting fraction if possible. If the denominator contains an irrational number (a radical), it is standard practice to rationalize the denominator.

Answer:

Answer:

If the numerator or the denominator (or both) are not perfect squares (e.g., $\sqrt{\frac{2}{5}}$ or $\sqrt{\frac{3}{8}}$), the square root of the fraction will involve irrational numbers. The result is often simplified by simplifying the radicals in the numerator and denominator and then rationalizing the denominator if it still contains a radical.

Example: Simplify $\sqrt{\frac{3}{8}}$.

$\sqrt{\frac{3}{8}} = \frac{\sqrt{3}}{\sqrt{8}}$. Simplify $\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$.

So, $\sqrt{\frac{3}{8}} = \frac{\sqrt{3}}{2\sqrt{2}}$. The denominator $2\sqrt{2}$ is irrational. Rationalize by multiplying numerator and denominator by $\sqrt{2}$.

$\frac{\sqrt{3}}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{3 \times 2}}{2 \times (\sqrt{2})^2} = \frac{\sqrt{6}}{2 \times 2} = \frac{\sqrt{6}}{4}$.

The square root of $\frac{3}{8}$ is $\frac{\sqrt{6}}{4}$.