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number system
Real
Number:A
real number
may be
either
rational or
irrational;
either
algebraic or
transcenden
tal; and
either
positive,
negative, or
zero
RATIONAL NUMBER
All terminating decimal are rational number
All non-terminating but repeating decimals are rational
number
All integer that can be expressed in the form of
p/q(where p and q are integer and q ≠ 0 is called
rational number.
Rational Number on Number line
Rational Number by successive
magnification(ex: 2.65)
REPRESENTATION OF 5.378
Irrational number on Number line
Geometrical representation of Real
Number
To proof square root3 is an irrational
number.
Solution:
To prove that this statement is true, let us assume that square root 3 is rational so
that we may write
Square root 3 = a/b
Here a and b = any two integers. We must then show that no two such integers can
be found.
Squaring both side
3 = a²/b²
3b² = a²
If b is odd then b² is odd. Similarly, if b is even, then b², a², and a are even. Since any
choice of even values of a and b leads to a ratio a/b that can be reduced by canceling
a common factor of 2.
Suppose a² is odd than then b is odd that is a=2m+1 and b=2n+1
Putting the value of a and b in the above equation
3(4n² + 4n + 1) = 4m² + 4m + 1
6n² + 6n + 1 = 2(m² + m)
The LHS of the above expression is odd and the RHS is even. That is a contradiction.
That is square root 3 is irrational.
Prove
square
root 2 is
irrational

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Number system

  • 2. Real Number:A real number may be either rational or irrational; either algebraic or transcenden tal; and either positive, negative, or zero
  • 3.
  • 4.
  • 5. RATIONAL NUMBER All terminating decimal are rational number All non-terminating but repeating decimals are rational number All integer that can be expressed in the form of p/q(where p and q are integer and q ≠ 0 is called rational number.
  • 6. Rational Number on Number line
  • 7. Rational Number by successive magnification(ex: 2.65)
  • 9. Irrational number on Number line
  • 11. To proof square root3 is an irrational number. Solution: To prove that this statement is true, let us assume that square root 3 is rational so that we may write Square root 3 = a/b Here a and b = any two integers. We must then show that no two such integers can be found. Squaring both side 3 = a²/b² 3b² = a² If b is odd then b² is odd. Similarly, if b is even, then b², a², and a are even. Since any choice of even values of a and b leads to a ratio a/b that can be reduced by canceling a common factor of 2. Suppose a² is odd than then b is odd that is a=2m+1 and b=2n+1 Putting the value of a and b in the above equation 3(4n² + 4n + 1) = 4m² + 4m + 1 6n² + 6n + 1 = 2(m² + m) The LHS of the above expression is odd and the RHS is even. That is a contradiction. That is square root 3 is irrational.