Step-by-step explanation:
√2+√3
Prove that √2 + √3 is an irrational number?
Let us assume that √2+√3 is a rational number.
It is in the form of p/q
√2+√3 = a/b
Where a and b are co - primes
=>√3 = (a/b) + √2
On squaring both sides, then
=>(√3)^2 = [(a/b/+√2]^2
This is in the form of (a+b)^2
Where a = (a/b) and b = √2
We know that (a+b)^2 = a^2+2ab+b^2
=>3 = (a/b)^2 + 2(a/b)(√2) + (√2)^2
=>3 = (a/b)^2 + 2√2a/b + 2
=>3-2 = (a^2/b^2 )+(2√2a/b)
=>1 = (a^2+2√2ab)/b^2
=>1×b^2 = a^2 + 2√2 ab
=>b^2 = a^2 + 2√2 ab
=> b^2-a^2=2√2ab
=>(b^2-a^2)/2ab = √2
=>√2 = (b^2-a^2)/(2ab)
=>√2 is in the form of p/q
=>√2 is not a raational number
This is a contradiction to our assumption.
=>√2 is an irrational number
Therefore,√2+√3 is an irrational number.
Hence,Proved.
Method of Contradiction or Indirect method is used for proving the given number is an irrational number.
The sum of two irrational numbers is also an irrational number.
So,
On squaring both sides, we get
Now,
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Verified answer
Step-by-step explanation:
Given:-
√2+√3
Required To Prove:-
Prove that √2 + √3 is an irrational number?
Proof:-
Let us assume that √2+√3 is a rational number.
It is in the form of p/q
√2+√3 = a/b
Where a and b are co - primes
=>√3 = (a/b) + √2
On squaring both sides, then
=>(√3)^2 = [(a/b/+√2]^2
This is in the form of (a+b)^2
Where a = (a/b) and b = √2
We know that (a+b)^2 = a^2+2ab+b^2
=>3 = (a/b)^2 + 2(a/b)(√2) + (√2)^2
=>3 = (a/b)^2 + 2√2a/b + 2
=>3-2 = (a^2/b^2 )+(2√2a/b)
=>1 = (a^2+2√2ab)/b^2
=>1×b^2 = a^2 + 2√2 ab
=>b^2 = a^2 + 2√2 ab
=> b^2-a^2=2√2ab
=>(b^2-a^2)/2ab = √2
=>√2 = (b^2-a^2)/(2ab)
=>√2 is in the form of p/q
=>√2 is not a raational number
This is a contradiction to our assumption.
=>√2 is an irrational number
Therefore,√2+√3 is an irrational number.
Hence,Proved.
Used Method:-
Method of Contradiction or Indirect method is used for proving the given number is an irrational number.
Note:-
The sum of two irrational numbers is also an irrational number.
So,
On squaring both sides, we get
Now,
So,