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May 9th, 2016, 11:04 PM   #1
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Talking recognizing one-to-one functions

Dear math people,

I am refreshing my brain cuz i'm planning to retake the equivalent of math 12 (precalc?) this summer before I go back to school and take math... can't remember ever studying this though... so right now I'm doing some self-study...

I am supposed to:

determine which of the given relations ins a one-to-one function over its domain.

{(1,a),(2,d),(3,b)}, {(1,b), (2,c),(3,b)}, {(1,a),(2,c),(3,b),(1,d)},
{(1,b),(1,c),(1,d),(2,a)}, {(1,d),(2,d),(3,d)}

I'm supposed to pick one. I wasn't sure how to answer this so I checked the answers at the back of the book and according to the back the correct answer is...

the back of the book stated that the correct answer is the very first set {(1,a),(2,d),(3,b)}, because there is only 1 x value to 1 y value and vice versa, but I'm not sure how they came to that conclusion from staring at a bunch of numbers/letters. As in, without graphing it out...

so effectively the back of the book didn't 'show their work' so I don't know how they came to that conclusion. My math 12 is rusty, please forgive me, for those of you that read this, please care to explain?

Please and thank you!
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May 10th, 2016, 12:48 AM   #2
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Try to draw a diagram for each of these. Write down the domains and co-domains in two separate columns, and line up the elements according to the binary relations. Then you'll see why.
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May 10th, 2016, 04:38 AM   #3
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Let's go over a couple of the incorrect choices ...

{(1,b), (2,c),(3,b)}, note the two function values, f(1) = f(3) = b, are the same value.
one-to-one functions have unique function values for every value in the domain

{(1,a),(2,c),(3,b),(1,d)}, not a function ... 1 maps to both a and d

Given the two explanations above ...

{(1,b),(1,c),(1,d),(2,a)}, you tell me what's wrong with this choice

{(1,d),(2,d),(3,d)}, you tell me what's wrong with this choice, also.
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May 10th, 2016, 04:19 PM   #4
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ohh i think i'm starting to get it...

{(1,b),(1,c),(1,d),(2,a)} - they are stating that the value of x has 3 different y values, (b c and d)... therefore it's not a one-to-one function because a one-to-one function has to have one x value to one y value and vice versa

{(1,d),(2,d),(3,d)} - similar idea, for the y value of 'd', there are 3 x values (1 and 2 and 3) meaning that it is also not a one-to-one function

i think i understand this particular question now thank you
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May 10th, 2016, 04:21 PM   #5
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lol it's been a long while since math 12
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May 10th, 2016, 04:31 PM   #6
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Quote:
{(1,b),(1,c),(1,d),(2,a)} - they are stating that the value of x has 3 different y values, (b c and d)... therefore it's not a one-to-one function because a one-to-one function has to have one x value to one y value and vice versa
it's just not a function
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