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.

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.

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:

• Room dimensions: 10m x 7m

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.

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.

are irrational numbers.

To show that 2+22+2

is irrational.

). Therefore, this example demonstrates that the product of a rational number and an irrational number can indeed be rational.