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Consider the function

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Answer #1

This is my attempt:

First, let x be rational; then f(x)=1/q is also rational. I is dense in R, and following the same argument as before, we can construct a sequence of irrational numbers {a_n} that converges to x, and so limn→∞f(a_n)=0≠f(x), meaning that f is not continuous for rational x.

Second, let x be irrational. Then f(x)=0. Let ϵ>0 and consider the inequality |a−x|<δ. We wish to show that choosing δ appropriately will mean that |f(a)−f(x)|=|f(a)|<ϵ. Suppose that aa is irrational; then f(a)=0 and δ can be chosen arbitrarily because |f(a)|=0<ϵ no matter what. Suppose, alternatively, that a is rational; let a=p/q where q>0 and p,q are relatively prime. Then f(a)=1/q, so we must choose δ so that 1/q<ϵ given |p/q−x|<δ. (Unsure how to complete this part.)

Third, we see that f(x) for nonzero x is either 0 or x/p for a nonzero integer p, meaning that |f(x)−f(0)|=|f(x)|<|x|. Let ϵ0 and choose δ=ϵ; then when |x|<δ we have |f(x)|<δ=ϵ. This demonstrates that ff is continuous at x=0.

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Answer 1

This is my attempt:

First, let x be rational; then f(x)=1/q is also rational. I is dense in R, and following the same argument as before, we can construct a sequence of irrational numbers {a_n} that converges to x, and so limn→∞f(a_n)=0≠f(x), meaning that f is not continuous for rational x.

Second, let x be irrational. Then f(x)=0. Let ϵ>0 and consider the inequality |a−x|<δ. We wish to show that choosing δ appropriately will mean that |f(a)−f(x)|=|f(a)|<ϵ. Suppose that aa is irrational; then f(a)=0 and δ can be chosen arbitrarily because |f(a)|=0<ϵ no matter what. Suppose, alternatively, that a is rational; let a=p/q where q>0 and p,q are relatively prime. Then f(a)=1/q, so we must choose δ so that 1/q<ϵ given |p/q−x|<δ. (Unsure how to complete this part.)

Third, we see that f(x) for nonzero x is either 0 or x/p for a nonzero integer p, meaning that |f(x)−f(0)|=|f(x)|<|x|. Let ϵ0 and choose δ=ϵ; then when |x|<δ we have |f(x)|<δ=ϵ. This demonstrates that ff is continuous at x=0.