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Assume it's rational. Then 2 + root2 = some rational number q. Then root2 = q - 2. However, the rational numbers are well-defined under addition by (a,b) + (c,d) = (ad + bc, bd) (in other words, you can add two fractions a/b and c/d and always get another fraction of the form (ad + bc)/bd.) Therefore, q - 2 = q + (-2) is rational, since both q and -2 are rational. This implies root2 must be rational, which is a contradiction. Therefore the assumption that 2 + root2 is rational must be false.
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Is there a proof to show that if you multiply an irrational number times an irrational number you get a rational number?
There cannot be a proof since your assertion is not necessarily true. sqrt(2)*sqrt(3) = sqrt(6). All three are irrational numbers.
How can the pythagorean spiral be used to show the exact value of an irrational number?
It cannot. It can only show square roots which represent only a small proportion of irrational numbers.
How do you show your work for the square root of 32 round to the nearest hundredths?
It is an irrational number and rounds to 5.66 to the nearest hundredths
How to prove that Assuming 5 is irrational how would you prove that 5 is irrational?
Root signs didn't show up
If you take the square root of an irrational number.Will it be irrational too?
Certainly. Otherwise, there would be a rational number whose square was an irrational number; that is not possible. To show this, let p/q be any rational number, where p and q are integers. Then, the square of p/q is (p^2)/(q^2). Since p^2 and q^2 must both be integers, their quotient is, by definition, a rational number. Thus, the square of every rational number is itself rational.
