Math, asked by nehaljaswal33, 11 months ago

Prove that 3+√7 is an irrational number​

Answers

Answered by Anonymous
1

Answer:

Here is ur answer

Step-by-step explanation:

Assume rt 7 is rational

rt 7 = a/b

in reduced form

Squarificate it

(rt7)^2 = (a/b)^2

7 = a^2 / b^2

a^2 = 7 b^2

But if a^2 is a multiple of 7 then it is a multiple of 49, and 'a' is a multiple of 7

a = 7c

(7c)^2 = 7 b^2

49c^2 = 7b^2

7c^2 = b^2

Same argument as before with 'a' being a multiple of 7, we see that b is a multiple of 7 now.

But that means a/b = 7c/ 7d = c/d is further reduced, thus contradicting our premise.

Conclusion...

rt7 is irrational.

3 + rt 7 is therefore also irrational

I hope it will help you a lot

Answered by Amazonalexa
0

ANSWER

We have to prove that 3+

7

is irrational.

Let us assume the opposite, that 3+

7

is rational.

Hence 3+

7

can be written in the form

b

a

where a and b are co-prime and b

=0

Hence 3+

7

=

b

a

7

=

b

a

−3

7

=

b

a−3b

where

7

is irrational and

b

a−3b

is rational.

Since,rational

= irrational.

This is a contradiction.

∴ Our assumption is incorrect.

Hence 3+

7

is irrational.

Hence proved.

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