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real numbers-

1-irrational numbers

A number which cannot be written in the form pqpq where pp and qq are integers and q≠0q≠0 is, Irrational Number are called. For example – 22, 33, , 0.101101110 . , , , , etc. are irrational numbers.

Theorem 1.3: Let’s say pp is a prime the number is. If pp divides a2a2, then pp will also divide aa where aa is a positive integer.

Proof:

real number

Let the prime factors of aa be of the following form:

a=p1,p2a=p1,p2, . , , pnpn where p2,p2p2,p2, . , , , , pnpn are prime numbers, but not necessarily differentiable.

Thus, a2=(p1P2…pn)(p1P2…pn)a2=(p1P2…pn)(p1P2…pn)

=p21p22…p2n=p12p22…pn2

Now given that pp divides a2a2. So, according to the Fundamental Theorem of Arithmetic; pp is a prime factor of a2a2. But using the property of uniqueness of the Fundamental Theorem of Arithmetic, we find that the prime factors of a2a2 are only p1,p2,…,pnp1,p2,…,pn.

Hence, pp must be from among p1,p1,…,pnp1,p1,…,pn.

Now, since a=p1p2…pna=p1p2…pn,

Hence, pp must divide aa.

Theorem 1.4 : 22 is an irrational number.

proof,

On the contrary we assume that 22 is a rational number.

So we can find two integers rr and ss such that 2=rs2=rs and s(≠0)s(≠0)

Let rr and ss have a common factor other than 1. Then we can divide rr and ss by this common factor to get 2=ab2=ab, where aa and bb are co-prime.

real number

So, b√2=ab2=a happened.

On squaring and rearranging both the sides, we get,

2b2=a22b2=a2 —–(i).

Hence 2 divides a2a2.

So we, by Theorem 1.3, will divide aa by 2 .

Thus, a=2ca=2c can be written, where cc is any integer.

Substituting a=2ca=2c in equation (i), we get

2b2=4c22b2=4c2

b2=2c2⇒b2=2c2

That is, 2 divides b2b2 and hence 2 will also divide bb.

Thus, aa and bb have at least one common factor of 2.

But this contradicts the fact that aa and bb have no common factors other than 1.

We have got this contradiction because we have mistakenly assumed that 22 is a rational number.

Hence, 22 is an irrational number.

real number

Some properties of 2-rational numbers:

The sum or difference of a rational number and an irrational number is an irrational number, and

The product or quotient of a non-zero rational number and an irrational number is an irrational number.

3-Some important questions of this questionnaire-

Question 1: Use Euclid’s division algorithm to find the HCF of the following numbers.

Question(a): 135 and 225

Answer: Let 225 = a and 135 = b

Here we use this equation in which a=bq+ra=bq+r

where r0

then get the following equation;

225=135×1+90225=135×1+90 where r=90r=90

Now for using Euclid’s division algorithm, let 135 = a and 90 = b

then get the following equation;

135=90×1+45135=90×1+45 where r=45r=45

Now for using Euclid’s division algorithm, suppose 90 = a and 45 = b

then get the following equation;

90=45×2+090=45×2+0

Here r = 0 is found.

Hence HCF = 45

Question 2-: 196 and 38220

Answer: Let 38220 = a and 196 = b

then get the following equation;

38220=196×195+038220=196×195+0

Here r=0r=0 is found.

Hence HCF = 196

Question 3- 867 and 255

Answer: Let 867 = a and 255 = b

So the following equation is obtained;

867=255×3+102867=255×3+102 where r=102r=102.

Now suppose 255 = a and 102 = b

So the following equation is obtained;

255=102×2+51255=102×2+51 where r=51r=51.

Now suppose 102 = a and 51 = b.

So the following equation is obtained.

102=51×2+0102=51×2+0

Here r = 0 is found.

Hence HCF = 51

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