show that root 2 is irrational
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Heya !!!
If possible, let ✓3 be rational Number and let it's simplest form be a/b.
Then, a and b are integers having no common factor other than 1 and b not equal to 0.
Now,
✓2 = a/b
2 = a²/b² [ On squaring both sides ]
=> 2b² = a² --------(1)
=> 2 divides a² [ Since 2 divides 2b². ]
=> 2 divides a
Let A = 2c for some integer C.
Putting A = 2c in equation (1) , we get
2b² = a²
2b² = (2c)²
2b² = 4c²
b² = 2c²
=> 2 divides b² [ 2 divides 2c² ]
=> 2 divides b.
Thus , 2 is a common factor of a and b.
But , this contradicts the fact that a and b have no common factor other than 1.
The contraction arises by assuming that ✓2 is rational number.
Hence,
✓2 is irrational number..... PROVED.....
★ HOPE IT WILL HELP YOU ★
If possible, let ✓3 be rational Number and let it's simplest form be a/b.
Then, a and b are integers having no common factor other than 1 and b not equal to 0.
Now,
✓2 = a/b
2 = a²/b² [ On squaring both sides ]
=> 2b² = a² --------(1)
=> 2 divides a² [ Since 2 divides 2b². ]
=> 2 divides a
Let A = 2c for some integer C.
Putting A = 2c in equation (1) , we get
2b² = a²
2b² = (2c)²
2b² = 4c²
b² = 2c²
=> 2 divides b² [ 2 divides 2c² ]
=> 2 divides b.
Thus , 2 is a common factor of a and b.
But , this contradicts the fact that a and b have no common factor other than 1.
The contraction arises by assuming that ✓2 is rational number.
Hence,
✓2 is irrational number..... PROVED.....
★ HOPE IT WILL HELP YOU ★
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