Learn UNIT I: Number systems with Free Lessons & Tips

To find the Highest Common Factor (HCF) of 52 and 117 and express it in the form 52x + 117y, we can use the Euclidean algorithm.

1. Step 1: Find the Remainder

   Divide the larger number by the smaller number and find the remainder.

   117=2×52+13117=2×52+13

   So, the remainder is 13.

2. Step 2: Replace Numbers

   Now, replace the divisor with the previous remainder, and the dividend with the divisor.

   52=4×13+052=4×13+0

   Here, the remainder is 0, so we stop. The divisor at this step, which is 13, is the HCF.

3. Expressing in the given form

   Now, we backtrack to express the HCF, which is 13, in the form 52x + 117y.

   Using the reverse steps of the Euclidean algorithm, we can express the HCF as a linear combination of 52 and 117:

   13=117−2×5213=117−2×52

   Hence, the HCF of 52 and 117 expressed in the form 52x + 117y is 13=117−2×5213=117−2×52.

To prove that x2−xx2−x is divisible by 2 for all positive integer xx, we can use mathematical induction.

Base Case: Let's start by checking the base case. When x=1x=1, x2−x=12−1=0x2−x=12−1=0. Since 0 is divisible by 2, the base case holds.

Inductive Hypothesis: Assume that x2−xx2−x is divisible by 2 for some positive integer kk, i.e., k2−kk2−k is divisible by 2.

Inductive Step: We need to show that if the statement holds for kk, then it also holds for k+1k+1.

(k+1)2−(k+1)=k2+2k+1−k−1=k2+k=(k2−k)+k(k+1)2−(k+1)=k2+2k+1−k−1=k2+k=(k2−k)+k

By the inductive hypothesis, we know that k2−kk2−k is divisible by 2. And we know that kk is a positive integer, so kk is also divisible by 2 or it is an odd number.

• If kk is divisible by 2, then k2−kk2−k is divisible by 2, and kk is divisible by 2, so their sum (k2−k)+k(k2−k)+k is also divisible by 2.

• If kk is an odd number, then k2−kk2−k is still divisible by 2, and when you add kk to it, you still get an even number, which is divisible by 2.

In both cases, (k+1)2−(k+1)(k+1)2−(k+1) is divisible by 2.

Therefore, by mathematical induction, we conclude that x2−xx2−x is divisible by 2 for all positive integers xx.

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:

• If kk and ll are both even, then m2+n2m2+n2 will leave a remainder of 22 when divided by 44.
• If kk and ll are both odd, then m2+n2m2+n2 will also leave a remainder of 22 when divided by 44.

Thus, m2+n2m2+n2 is even but not divisible by 44 when both mm and nn are odd positive integers.

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:

1. 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.

2. 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.

3. 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.

4. Therefore, n divides a.

So, the statement is justified by the properties of prime numbers and the Fundamental Theorem of Arithmetic.

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.

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:

1. x≡8(mod28)x≡8(mod28)
2. x≡12(mod32)x≡12(mod32)

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.

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:

1. 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.

2. 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.

3. 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.

Sure, let's prove this by induction.

Base Case:
When n=2n=2, a2=a×aa2=a×a, where aa is a positive rational number. The product of two rational numbers is rational, so a2a2 is rational.

Inductive Step:
Assume that anan is rational for some positive integer n>1n>1.

Consider an+1an+1: an+1=an×aan+1=an×a

By the inductive hypothesis, anan is rational. And since aa is rational (given in the problem), the product of two rational numbers is rational. Therefore, an+1an+1 is rational.

By mathematical induction, anan is rational for all positive integers n>1n>1.

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