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Post a LessonAnswered on 17 Apr Learn CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/Euclid's Division Lemma
Nazia Khanum
To find the size of tiles required and how many tiles are needed, we first need to determine the largest possible size of the square marble tiles that can fit evenly into the room's dimensions.
Given:
To find the largest possible size of the square tiles, we need to find the greatest common divisor (GCD) of the dimensions (length and width) of the room.
GCD(10, 7) = 1
This means that there's no integer length that evenly divides both 10 and 7. So, we cannot find a single tile size that perfectly fits the room without cutting.
However, we can approximate the largest possible tile size by using factors of the GCD, which in this case is 1. So, theoretically, we could use a tile size of 1m x 1m, but that wouldn't be practical.
In real-world scenarios, the tile size is usually chosen for convenience and aesthetics. A common approach is to use a tile size that evenly divides the room dimensions, even if it involves some cutting.
Let's say we decide to use a 0.5m x 0.5m tile size. Then, we can calculate how many tiles are needed:
For the length of the room (10m), we would need 10m / 0.5m = 20 tiles. For the width of the room (7m), we would need 7m / 0.5m = 14 tiles.
So, in total, we would need 20 tiles lengthwise and 14 tiles widthwise, resulting in 20 x 14 = 280 tiles.
However, keep in mind that some tiles will need to be cut to fit the edges of the room, especially along one of the dimensions (either length or width). The number of tiles that need to be cut will depend on the exact layout of the tiles and the dimensions of the room.
Answered on 17 Apr Learn CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/Euclid's Division Lemma
Nazia Khanum
This statement is a direct consequence of a fundamental property in number theory known as the "Fundamental Theorem of Arithmetic" and some basic properties of prime numbers.
The Fundamental Theorem of Arithmetic states that every integer greater than 1 either is a prime number itself or can be represented as the product of prime numbers, and this representation is unique, up to the order of the factors. In other words, any integer greater than 1 can be expressed as a unique product of prime numbers.
Now, let's consider the given statement:
"If n is any prime number and a^2 is divisible by n, then n will also divide a."
Proof:
Let's assume that n is a prime number, and a2a2 is divisible by n. This implies that a2=kna2=kn, where k is some integer.
According to the Fundamental Theorem of Arithmetic, a2a2 can be expressed as the product of prime factors. Since n is prime, it must be one of the prime factors of a2a2.
If n is a factor of a2a2, then n must also be a factor of a (this follows from the uniqueness of prime factorization). This is because if a2=kna2=kn, then a must contain at least one factor of n, as otherwise, a2a2 would not be divisible by n.
Therefore, n divides a.
So, the statement is justified by the properties of prime numbers and the Fundamental Theorem of Arithmetic.
Answered on 26/11/2022 Learn CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/Euclid's Division Lemma
Sandhya
Maths tutor with 7 years experience
Answered on 17 Apr Learn CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/Rational and irrational numbers
Nazia Khanum
Sure, let's consider the rational number 1221 and the irrational number 22
.
The product of 1221 and 22
is:
12×2=2221×2
=22
Here, we have a rational number (1221) multiplied by an irrational number (22
), resulting in another rational number (2222
). Therefore, this example demonstrates that the product of a rational number and an irrational number can indeed be rational.
Answered on 17 Apr Learn CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/Rational and irrational numbers
Nazia Khanum
To prove that √3 – √2 and √3 + √5 are irrational, we can use proof by contradiction.
So, 3−2=ab3
−2
=ba.
Squaring both sides, we get: 3−26+2=a2b23−26
+2=b2a2 ⇒6=a2−12b2⇒6
=2b2a2−1
This implies 66
is rational. However, we know that 66 is irrational (since 6 is not a perfect square), which contradicts our assumption. Thus, 3−23−2
must be irrational.
+5
So, 3+5=cd3
+5
=dc.
Squaring both sides, we get: 3+215+5=c2d23+215
+5=d2c2 ⇒15=c2−8d24d2⇒15
=4d2c2−8d2
This implies 1515
is rational. However, we know that 1515 is irrational (since 15 is not a perfect square), which contradicts our assumption. Thus, 3+53+5
must be irrational.
Therefore, both 3−23
−2 and 3+53+5
are irrational.
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