prove-by-contradiction-9-13-3-is-irrational-

Question Number 12822 by fawadalamawan@gmail.com last updated on 02/May/17

p r o v e b y c o n t r a d i c t i o n 9 + 13 \sqrt{3}

i s i r r a t i o n a l

Answered by mrW1 last updated on 03/May/17

l e t u s a s s u m e 9 + 13 \sqrt{3} i s r a t i o n a l, . i . e .

t h e r e e x i s t i n t e g e r n u m b e r s a a n d b, b \neq 0,

9 + 13 \sqrt{3} = \frac{a}{b}

\Rightarrow \sqrt{3} = \frac{a - 9 b}{13} = \frac{i n t e g e r}{i n t e g e r} = r a t i o n a l

t h a t i s t o s a y : i f 9 + 13 \sqrt{3} i s a s s u m e d

t o b e r a t i o n a l, t h e n \sqrt{3} i s a l s o r a t i o n a l,

b u t t h i s i s n o t t r u e, t h e r e f o r e

9 + 13 \sqrt{3} i s i r r a t i o n a l .

Commented by Joel577 last updated on 04/May/17

h o w t o p r o v e t h a t \sqrt{3} i s i r r a t i o n a l ?

Commented by mrW1 last updated on 04/May/17

a s s u m e \sqrt{3} w e r e r a t i o n a l, i . e .

\sqrt{3} = \frac{a}{b}, a n d \frac{a}{b} c a n n o t b e f u r t h e r

s i m p l i f i e d, i . e . a a n d b e h a v e n o

c o m m o n f a c t o r .

3 = \frac{a^{2}}{b^{2}}

a^{2} = 3 b^{2}

i f b i s e v e n, i . e . b = 2 n

a^{2} = 3 \times 4 n^{2} = e v e n

\Rightarrow a = e v e n = 2 m

\Rightarrow a a n d b h a v e c o m m o n f a c t o r 2,

\frac{a}{b} c a n b e f u r t h e r s i m p l i f i e d t o \frac{m}{n},

\Rightarrow b i s n o t e v e n!

i f b i s o d d, i . e . b = 2 n + 1

b^{2} i s a l s o o d d, 3 b^{2} i s a l s o o d d, i . e .

a^{2} i s o d d, a m u s t b e a l s o a d d, i . e . a = 2 m + 1

{(2 m + 1)}^{2} = 3 {(2 n + 1)}^{2}

4 m^{2} + 4 m + 1 = 12 n^{2} + 12 n + 3

2 (m^{2} + m) = 6 (n^{2} + n) + 1

e v e n = o d d

\Rightarrow b i s n o t o d d .

\Rightarrow t h e r e e x i s t n o i n t e g e r a a n d b t o f u l f i l l \sqrt{3} = \frac{a}{b},

\Rightarrow \sqrt{3} i s i r r a t i o n a l

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