Square roots of numbers that are not the squares of integers are all irrational. There is a way to prove that but I forgot the details. You have to assume that a fraction works and then prove that the assumption leads to a contradiction
Any number that does not meet the definition of being rational is irrational.
- √2 is either rational or irrational.
- Assume that √ is rational so that
√2 = a/b, with a/b in lowest terms
- Square both sides to get
2 = a^2/b^2
then a^2 = 2b^2
- the right side of this eqation is clearly an even number, since anything multiplied by 2 is even
- so a^2 must be even. We also know that if we square an odd number the result is odd, and if we square an even number the result is even
so a must be even
so a could be written as 2k
- rewriting our equation as
2b^2 = 4k^2
b^2 = 2k^2
by the same argument as above 2k^2 is even , so b has to be even
which means a and b are both even, therefore a/b is not in lowest terms
BUT that contradicts my assumption that a/b was a fraction in lowest terms
so √2 = a/b is a false statement
therefore √2 cannot be rational, and
must then be irrational
the same argument could be used for √3 and all other square roots
true or false does this number represent a rational number 0.20200200020000200000
Is √4 Irrational?
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