Prove that 2√2-1 is an irrational number
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Answered by
11
Hey.. I think this can be ur answer!!
To prove: 2root2 - 1 irrational.
Let us assume that 2root2 - 1 is rational so,
2root2 - 1 = a/b (a and b are co prime)
2root2 = a/b+1 = a+b/b
Now,
Root2 = a+b/2b
BUT we know that root2 is irrational
So 2root2 - 1 is also irrational
Hence, our assumption was wrong.
Hence proved!!
Hope it helps you dear!!
To prove: 2root2 - 1 irrational.
Let us assume that 2root2 - 1 is rational so,
2root2 - 1 = a/b (a and b are co prime)
2root2 = a/b+1 = a+b/b
Now,
Root2 = a+b/2b
BUT we know that root2 is irrational
So 2root2 - 1 is also irrational
Hence, our assumption was wrong.
Hence proved!!
Hope it helps you dear!!
aryaniron2003:
Thanks a lot
Answered by
7
let be 2√2-1 is an rational number.
then their exist a and b positive integer.
now
2√2-1 = a/b ( where a and b are co primes)
2√2=a/b +1
2√2= a+b/b
√2= a+b/2b ( by transposing LHS to rhs )
but we know √2 is an irrational no.
and a and b are positive integer so,
a+b/2b is an rational no.
this is a contradiction
because LHS is always equal to RHS.
therefore our assumption is wrong.
hence , 2√2-1 is an irrational number.
then their exist a and b positive integer.
now
2√2-1 = a/b ( where a and b are co primes)
2√2=a/b +1
2√2= a+b/b
√2= a+b/2b ( by transposing LHS to rhs )
but we know √2 is an irrational no.
and a and b are positive integer so,
a+b/2b is an rational no.
this is a contradiction
because LHS is always equal to RHS.
therefore our assumption is wrong.
hence , 2√2-1 is an irrational number.
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