Question

prove root 3is irrational . hence show that 7➕root 3 is irrational

Answer 1

irrational number are those whose decimal nature behave non terminating.Hence value of root 3 is 1.738..........
so it is an irrational number.
And when rational number is added to an irrational number then the number formed is irrational.So 7 plus root 3 is also an irrational number.

Answer 2

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rishu6367 avatar

rishu6367

02.03.2019

Math

Secondary School

+5 pts

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Prove that root 3 is an irrational number hence show that 7 + 2 root 3 is also an irrational number

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nehaputhan

NehaputhanAmbitious

Hey

√3 is irrational.

This can be proved by the method of contradiction. Let us assume that √3 is rational.

So we can find co-primes a and b such that √3 = a/b and b not equal to 0

that is       b√3 = a

Squaring, (b√3)² = a²

      or,       a² = 3b² _________(1)

ie, a² is divisible by 3 or 3 divides a²  

Therefore, 3 divides a also ie, 3 is a factor of a . Since 3 is a factor of a , we can find another integer c such that

 a = 3c

Squaring, a² = (3c)² = 9c²

                3b² = 9c² (from eq. (1) )

                  b² = 3c²

ie, b² is also a multiple of 3 or 3 divides b² and hence 3 divides b also

That is 3 is a common factor of a and b contradicting the fact that a and b are co-primes .

This contradiction arises due to the wrong assumption that √3 is rational. Hence √3 is irrational.

Now we have to prove that 7 + 2√3 is irrational. This can also be proved by the method of contradiction.

Let us assume that 7 + 2√3 is rational. Therefore we can find co-primes a and b such that 7 + 2√3 = a/b where b not equal to 0

                                 2√3 = a/b - 7 = a-7b/b

                                    √3 = a-7b/2b which is rational

ie , √3 is rational , which is a contradiction to the fact that √3 is irrational.

This contradiction arises due to the fact the wrong assumption that 7 + 2√3 is rational.

Hence it is proved that 7 + 2√3 is irrational.

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rishu6367

and which class do you read

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Janani7305Expert

Hey

We know that √3 is irrational.This can be proved by the method of contradiction.

Let us assume that √3 is rational.

So we can find co-primes a and b such that √3 = a/b and b not equal to 0

that is =      b√3 = a

Squaring, (b√3)² = a²

      or,       a² = 3b² _________(1)

ie, a² is divisible by 3 or 3 divides a²  

Therefore, 3 divides a also ie, 3 is a factor of a . Since 3 is a factor of a , we can find another integer c such that

 a = 3c

Squaring, a² = (3c)² = 9c²

                3b² = 9c² (from eq. (1) )

                  b² = 3c²

ie, b² is also a multiple of 3 or 3 divides b² and hence 3 divides b also

That is 3 is a common factor of a and b contradicting the fact that a and b are co-primes .

This contradiction arises due to the wrong assumption that √3 is rational. Hence √3 is irrational.

Now we have to prove that 7 + 2√3 is irrational. This can also be proved by the method of contradiction.

Let us assume that 7 + 2√3 is rational. Therefore we can find co-primes a and b such that 7 + 2√3 = a/b where b not equal to 0

                                 2√3 = a/b - 7 = a-7b/b

                                    √3 = a-7b/2b which is rational

ie , √3 is rational , which is a contradiction to the fact that √3 is irrational.

This contradiction arises due to the fact the wrong assumption that

7 + 2√3 is rational.

Hence it is proved that 7 + 2√3 is irrational.

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