tatianagomezb
  • tatianagomezb
True or false: If a lineal function is injective, then it is subjective too. How would I explain why this is true? I know the difference between both.
Mathematics
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katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
I think you meant to say `linear` Any linear function that is not a constant function will be both injective and surjective. Combining the two means this linear function is bijective
jim_thompson5910
  • jim_thompson5910
any linear function is in the form f(x) = mx+b assume m is nonzero now let's say we have two inputs x1 and x2. The claim is that if f(x1) = f(x2) then x1 = x2 f(x) = mx+b f(x1) = mx1+b f(x2) = mx2+b setting up f(x1) = f(x2) leads to mx1 = mx2 which leads to x1 = x2. So this all proves that f(x) = mx+b is injective where m is nonzero
jim_thompson5910
  • jim_thompson5910
to prove f(x) = mx+b is surjective (m is still nonzero), you simply have to show that you can target any y value you want. Which effectively means that the range is the set of all real numbers hopefully that makes sense?

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tatianagomezb
  • tatianagomezb
I always forget is linear, lineal is in spanish. It makes more sense now, I'm having problems with the range, is it supposed to be equal to the domain or codomain. That's where I always make mistakes.
jim_thompson5910
  • jim_thompson5910
oh I did not know that about spanish. What I would do is solve y = mx+b for x doing so gives you x = (y-b)/m notice how m must be nonzero to avoid division by zero errors if m is nonzero, then we can solve y = mx+b for x which means that we can target any y value we want. Therefore this proves f(x) = mx+b, with m nonzero, is also surjective as well
tatianagomezb
  • tatianagomezb
I got it! Thanks a lot. Then this is a matter of mentioning the properties of the linear function: they are bijective, because they're injective and subjective; and eventually finding the inverse.

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