# 11x1 t02 02 rational & irrational (2011)

Post on 14-Jul-2015

439 views

Category:

## Education

Embed Size (px)

TRANSCRIPT

• Rational Numbers

• Rational NumbersRational numbers can be expressed in the form where a and b are integers. b

a

• Rational NumbersRational numbers can be expressed in the form where a and b are integers. b

a

Irrational Numbers

• Rational NumbersRational numbers can be expressed in the form where a and b are integers. b

a

Irrational NumbersIrrational numbers are numbers which are not rational.

All irrational numbers can be expressed as a unique infinite decimal.

• irrational is 2 Prove ..ge

• irrational is 2 Prove ..geProof by contradiction

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factors

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

222 ab

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

222 ab 2by divisible bemust Thus 2a

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

222 ab 2by divisible bemust Thus 2a

4by divisible is2by divisibleisthat squareAny

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

222 ab 2by divisible bemust Thus 2a

4by divisible is2by divisibleisthat squareAny 4by divisible bemust Thus 2a

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

222 ab 2by divisible bemust Thus 2a

4by divisible is2by divisibleisthat squareAny 4by divisible bemust Thus 2a

2 2 4b k where k is an integer

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

222 ab 2by divisible bemust Thus 2a

4by divisible is2by divisibleisthat squareAny 4by divisible bemust Thus 2a

2 2 4b k where k is an integer2 2b k

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

222 ab 2by divisible bemust Thus 2a

4by divisible is2by divisibleisthat squareAny 4by divisible bemust Thus 2a

2 2 4b k

factorcommon a havemust and 2by divisibleboth are and So 22 ba

where k is an integer2 2b k

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

222 ab 2by divisible bemust Thus 2a

4by divisible is2by divisibleisthat squareAny 4by divisible bemust Thus 2a

2 2 4b k

factorcommon a havemust and 2by divisibleboth are and So 22 baHowever, a and b have no common factors

where k is an integer2 2b k

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

222 ab 2by divisible bemust Thus 2a

4by divisible is2by divisibleisthat squareAny 4by divisible bemust Thus 2a

2 2 4b k

factorcommon a havemust and 2by divisibleboth are and So 22 baHowever, a and b have no common factors

rationalnot is 2so

where k is an integer2 2b k

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

222 ab 2by divisible bemust Thus 2a

4by divisible is2by divisibleisthat squareAny 4by divisible bemust Thus 2a

2 2 4b k

factorcommon a havemust and 2by divisibleboth are and So 22 baHowever, a and b have no common factors

rationalnot is 2soirrational is2

where k is an integer2 2b k

• irrational is 2 Prove ..geProof by contradiction

rationalis 2 Assume

ba 2 where a and b are integers with no common factorsab 2

222 ab 2by divisible bemust Thus 2a

4by divisible is2by divisibleisthat squareAny 4by divisible bemust Thus 2a

2 2 4b k

factorcommon a havemust and 2by divisibleboth are and So 22 baHowever, a and b have no common factors

rationalnot is 2soirrational is2 Exercise 2B; 2a, 3, 4 (root 3), 8, 14*

where k is an integer2 2b k

Slide Number 1Slide Number 2Slide Number 3Slide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20

Recommended