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2012年6月13日 · Corollary. $$\mathbb{Q} +\mathbb{Q}^c \subseteq \mathbb{Q}^c$$ That is: "A rational number plus an irrational number will always itself be irrational." The opening observation of user17762's answer is trickier.
2013年10月29日 · I proposed that 1 + an irrational number is always irrational, thus if I could prove that 1 + irrational number is irrational, then it stood to reason that was also proving that the number in question was irrational. Eg. √2 + 1 can be expressed as a continuous fraction, and through looking at the fraction, it can be assumed √2 + 1 is ...
2014年3月16日 · Prove that e is an irrational number. Recall that e = ∞ ∑ n = 0 1 n!, and assume e is rational, then. ∞ ∑ k = 0 1 k! = a b, for some positive integers a, b. so. b ∞ ∑ k = 0 1 k! = a. or. b(1 + 1 + 1 2 + 1 6 + ⋯) = a.
Possible Duplicate: Density of irrationals I am trying to prove that there exists an irrational number between any two real numbers a and b. I already know that a rational number between the t...
2014年1月3日 · An irrational number is one that cannot be written in the form. a b a b. where a a and b b are integers; a rational number is what that can be written in that form. Now it is the case that a number has a repeating base-10 representation if, and only if, it is a rational number, that is if it can be written as a fraction a b a b.
Claim: If x x is irrational and r ≠ 0 r ≠ 0 is rational, then xr x r is irrational. Proof: Suppose that xr x r were rational. Then, x = xr r x = x r r would be rational (as the quotient of two rationals). This clearly contradicts the assumption that x x is irrational. Therefore, xr x r is irrational.
Q: does the multiplication of a and b result in a rational or irrational number?: Proof: because b is rational: b = u/j where u and j are integers Assume ab is rational: ab = k/n, where k and n are integers. a = k/bn a = k/(n(u/j)) a = jk/un before we declared a as
Yes, for every rational number (with the exception of 0 itself, of course), there's an irrational number that's closer to 0. But there's no irrational number which is closer to 0 than every rational number, because for each irrational number there's a rational number which is even closer to 0. – celtschk.
Those couples of magnitudes was called "incommensurable" (i.e. without common measure). For the same reason, 2–√ 2 is an irrational number, exactly because the ratio "diagonal/side" is not expressible as a ratio between natural numbers. The irrationality of π π was proved by Johann Heinrich Lambert in 1761. Share.
6 = p2 q2. ⇒ 6q = p2 q. It is clear that the left hand side is an integer. But the right hand side isn't since p2 and q share no common factors. So this equality can not hold. And √6 can not equal p q. So it has to be an irrational number. There's an incredibly short proof of this if you know the rational root theorem.
