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🔶03 - Show that Square Root of 3 is irrational | Square Root of all Prime Numbers are Irrational
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🔶02 - Show that Square Root of 3 is irrational
In this video, we are going to prove that square root of 3 is irrational.
An irrational number is a number that cannot be expressed as a fraction as a/b , where b is not equal to 0
To solve this:
first lets assume that root 3 is rational, thus
root 3 = a/b,
we also assume that, root 3 is irreducible/simplified and a and b are co-prime, thus the only common factor between a and b is 1.
root 3 = a/b
3 = a2/b2
3b2 = a2..............(1)
here we have 3 times a certain number b2 = a2,
thus
3 is a factor of a2 and
a2 is said to be a multiple of 3,
now from theorem, if a2 is a multiple of k, then a is also a multiple of k, thus
a is also a multiple of 3, hence:
3 is a factor of a.
since 3 is a factor of a, it means a = 3 times a certain integer value say c,
then from equ1,
3b2 = (3c)2 = 9c2
3b2 = 9c2
b2 = 3c2, thus 3 is a factor of b2, and hence a factor of b,
there we have our contradiction, because we assumed initially that a and b were co-prime and have a common factor of 1, but here
a and b have a common factor of three hence the fraction is reducible so:
root 3 cannot be written as a fraction,
and hence
root 3 is irrational
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In this video, we are going to prove that square root of 3 is irrational.
An irrational number is a number that cannot be expressed as a fraction as a/b , where b is not equal to 0
To solve this:
first lets assume that root 3 is rational, thus
root 3 = a/b,
we also assume that, root 3 is irreducible/simplified and a and b are co-prime, thus the only common factor between a and b is 1.
root 3 = a/b
3 = a2/b2
3b2 = a2..............(1)
here we have 3 times a certain number b2 = a2,
thus
3 is a factor of a2 and
a2 is said to be a multiple of 3,
now from theorem, if a2 is a multiple of k, then a is also a multiple of k, thus
a is also a multiple of 3, hence:
3 is a factor of a.
since 3 is a factor of a, it means a = 3 times a certain integer value say c,
then from equ1,
3b2 = (3c)2 = 9c2
3b2 = 9c2
b2 = 3c2, thus 3 is a factor of b2, and hence a factor of b,
there we have our contradiction, because we assumed initially that a and b were co-prime and have a common factor of 1, but here
a and b have a common factor of three hence the fraction is reducible so:
root 3 cannot be written as a fraction,
and hence
root 3 is irrational
Visit channel Playlist for more videos on Engineering mathematics, applied electricity and Basic Mechanics.
Kindly support and Subscribe
Thank you
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