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Answered on 17 Apr Learn Euclid's Division Lemma
Nazia Khanum
Sure, let's prove this statement.
Let's start with m=2k+1m=2k+1 and n=2l+1n=2l+1, where kk and ll are integers.
Now, let's square both mm and nn:
m2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1m2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1 n2=(2l+1)2=4l2+4l+1=2(2l2+2l)+1n2=(2l+1)2=4l2+4l+1=2(2l2+2l)+1
Now, let's sum them:
m2+n2=2(2k2+2k)+1+2(2l2+2l)+1=2(2k2+2k+2l2+2l)+2m2+n2=2(2k2+2k)+1+2(2l2+2l)+1=2(2k2+2k+2l2+2l)+2
Since 2k2+2k+2l2+2l2k2+2k+2l2+2l is an integer, let's denote it as qq. Then:
m2+n2=2q+2m2+n2=2q+2
This clearly shows that m2+n2m2+n2 is even, as it is divisible by 22.
To prove that m2+n2m2+n2 is not divisible by 44, let's consider the possible remainders when dividing by 44:
Thus, m2+n2m2+n2 is even but not divisible by 44 when both mm and nn are odd positive integers.
Answered on 17 Apr Learn Euclid's Division Lemma
Nazia Khanum
To find aa, we can use the relationship between the highest common factor (HCF) and the least common multiple (LCM).
Given: HCF(6,a)=2HCF(6,a)=2 LCM(6,a)=60LCM(6,a)=60
We know that the product of the HCF and LCM of two numbers is equal to the product of the numbers themselves. So, we have:
HCF(6,a)×LCM(6,a)=6×aHCF(6,a)×LCM(6,a)=6×a
Substituting the given values:
2×60=6×a2×60=6×a 120=6a120=6a
Now, we solve for aa:
a=1206a=6120 a=20a=20
Therefore, a=20a=20.
Answered on 17 Apr Learn 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.
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Answered on 17 Apr Learn 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 17 Apr Learn Euclid's Division Lemma
Nazia Khanum
To find the smallest number that is divisible by both 90 and 144 when increased by 20, we need to find the least common multiple (LCM) of 90 and 144. Then, we'll add 20 to that LCM to get our answer.
First, let's find the LCM of 90 and 144.
The prime factorization of 90 is 2×32×52×32×5.
The prime factorization of 144 is 24×3224×32.
To find the LCM, we take the highest power of each prime factor that appears in either number:
LCM=24×32×5=720LCM=24×32×5=720
Now, we add 20 to 720:
720+20=740720+20=740
So, the smallest number that, when increased by 20, is exactly divisible by both 90 and 144 is 740.
Answered on 17 Apr Learn Euclid's Division Lemma
Nazia Khanum
To find the smallest number that leaves remainders of 8 when divided by 28 and 12 when divided by 32, we can use the Chinese Remainder Theorem (CRT).
The CRT states that if we have a system of congruences x≡ai(modmi)x≡ai(modmi) for i=1,2,…,ni=1,2,…,n, where the mimi are pairwise coprime, then there exists a unique solution xx modulo M=m1⋅m2⋅…⋅mnM=m1⋅m2⋅…⋅mn.
In our case, we have:
First, let's find the value of M=28×32=896M=28×32=896.
Next, we find the multiplicative inverses of 32 modulo 28 and of 28 modulo 32. Let's call these inverses y1y1 and y2y2 respectively.
y1y1 is such that 32×y1≡1(mod28)32×y1≡1(mod28). y2y2 is such that 28×y2≡1(mod32)28×y2≡1(mod32).
Using the Extended Euclidean Algorithm or observation, we find y1=22y1=22 and y2=9y2=9.
Now, we can use these inverses to find the solution:
x=(8×32×9+12×28×22)(mod896)x=(8×32×9+12×28×22)(mod896)
Let's compute this:
x=(2304+7392)(mod896)x=(2304+7392)(mod896) x=9696(mod896)x=9696(mod896) x=48x=48
So, the smallest number that satisfies the conditions is 48.
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Answered on 17 Apr Learn Rational and irrational numbers
Nazia Khanum
Euclid's division lemma states that for any two positive integers, aa and bb, there exist unique integers qq and rr such that:
a=bq+ra=bq+r
where 0≤r<b0≤r<b.
Let's prove the given statement using Euclid's division lemma.
Consider any positive integer nn. We want to show that n2n2 can be expressed in the form 3m3m or 3m+13m+1 for some integer mm.
First, let's divide nn by 33 using Euclid's division lemma:
n=3q+rn=3q+r
where 0≤r<30≤r<3.
Now, let's square both sides:
n2=(3q+r)2n2=(3q+r)2
Expanding the right side:
n2=9q2+6qr+r2n2=9q2+6qr+r2
Now, consider the possible values of rr:
If r=0r=0, then n2=9q2n2=9q2. Since 9q29q2 is divisible by 33 (because each term 9q29q2 is divisible by 33), we can express n2n2 in the form 3m3m, where m=3q2m=3q2.
If r=1r=1, then n2=9q2+6q+1=3(3q2+2q)+1n2=9q2+6q+1=3(3q2+2q)+1. Here, n2n2 is of the form 3m+13m+1, where m=3q2+2qm=3q2+2q.
If r=2r=2, then n2=9q2+12q+4=3(3q2+4q+1)+1n2=9q2+12q+4=3(3q2+4q+1)+1. Here, n2n2 is of the form 3m+13m+1, where m=3q2+4q+1m=3q2+4q+1.
In all cases, n2n2 is either of the form 3m3m or 3m+13m+1 for some integer mm, which completes the proof.
Answered on 17 Apr Learn Rational and irrational numbers
Nazia Khanum
To show that 2+22+2
is not a rational number, we'll use a proof by contradiction.
Assume that 2+22+2
is rational. That means it can be expressed as the ratio of two integers aa and bb where b≠0b=0 and aa and bb have no common factors other than 1:
2+2=ab2+2
=ba
Now, let's rearrange this equation to isolate 22
:
2=ab−22
=ba−2
2=a−2bb2
=ba−2b
Now, square both sides:
2=(a−2bb)22=(ba−2b)2
2=(a−2b)2b22=b2(a−2b)2
2b2=(a−2b)22b2=(a−2b)2
2b2=a2−4ab+4b22b2=a2−4ab+4b2
0=a2−4ab+2b20=a2−4ab+2b2
This equation represents a quadratic equation in terms of aa. Now, let's consider this equation modulo 2. This means we'll look at the remainders when dividing each term by 2.
0≡a2−4ab+2b2(mod2)0≡a2−4ab+2b2(mod2)
0≡a2(mod2)0≡a2(mod2)
Since the square of any integer is congruent to either 0 or 1 modulo 2, a2≡0(mod2)a2≡0(mod2) implies that aa itself must be even.
Let a=2ka=2k, where kk is an integer.
Now, substitute a=2ka=2k into the equation:
0=(2k)2−4(2k)b+2b20=(2k)2−4(2k)b+2b2
0=4k2−8kb+2b20=4k2−8kb+2b2
0=2(2k2−4kb+b2)0=2(2k2−4kb+b2)
Since 22 is a prime number, for 2(2k2−4kb+b2)2(2k2−4kb+b2) to be 00, the term inside the parentheses must be divisible by 22. But if 22 divides 2k2−4kb+b22k2−4kb+b2, then 22 divides each of its terms, including b2b2. This implies that bb is also even.
Now, if both aa and bb are even, then they have a common factor of 22, contradicting our initial assumption that aa and bb have no common factors other than 1.
Thus, our initial assumption that 2+22+2
is rational must be false. Therefore, 2+22+2
is irrational.
Answered on 17 Apr Learn 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.
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Answered on 17 Apr Learn 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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