what does 3b^2=a^2 mean??? (I am trying to prove that 3 is an irrational number and after some algebraic some manipulation got left with that.)
what are a and b here?? natural numbers,integers,real numbers?
natural numbers
would help if I showed you what I done before hand??
yea it would be great :D
I am trying to prove √3 is a irrational number by contradiction right, so √3 = a/b ( where a and b are simplified to lowest terms) Next I raised the equation to the ^2 to get 3 = a^2/b^2 Next I multiplied both sides of the equation by b^2 to get 3b^2=a^2.
Assume that sqrt3 is rational and equal to a/b in lowest terms with a and b integers. Squaring both sides gives 3 = a^2/b^2 so 3b^2 = a^2. If a^2 is a multiple of 3 and a is an integer then a itself must be a multiple of 3 (see question a few prior to this one). Since a is a multiple of 3 replace it by 3c. Then 3b^2 = (3c)^2 = 9c^2 which reduces to b^2 = 3c^2. Again b^2 a multiple of 3 means b itself is a multiple of 3. We have now shown that a, b are both multiples of 3 contradicting the statement that a/b was in lowest terms, i.e. no common factors. This proves that the original assumption that sqrt3 is rational must be false.
:D
this is how i had replied once to some1 who needed help in proving that.. hope it helps :)
you get it right?
I dont get why we have to substitute 3c in for a^2
as it is a multiple of 3.. say it was 9 so we substituted it as 3*3
so what relationship do 3c and a^2 have?
we have a=3c
and why is it 3c and not just 3?
see we have a^2 is a multiple of 3 right?? as it is 3b^2..so as a^2 is a multiple of 3 we have by implication even a is a multiple of 3... it is a multiple so we dont write just 3 but as 3c as a multiple can be any number other than 3 as well
ok great
so a rational number can be expressed as a fraction in lowest terms and a irrational number cant
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