Use a proof by cases to show that min(a, min(b,c)) = min(min(a,b),c) whenever a,b and c are real numbers.

Can anyone help please? I don't know where to start and what does min() mean?

Thanks

min(x,y) generally will return the minimum or smaller of the two numbers

min(a,min(b,c)) will return the smallest of the 3 and so will min(min(a,b),c))

Proof by cases means make up numbers and prove that it works multiple times.

hm i thought it means determine the different scenarios and show it's true for all of them.

eg:
1) a < b < c
2) a < b > c
3) a > b < c
4) a > b > c
then all the equal cases too

This post was edited by carteblanche on Oct 17 2015 09:30pm

last question:

For some n belonging to Z+, if n is odd then n^2=8m +1 for some m belonging to Z+ how do we prove this by case?

thank you

as a starting place, i'd expand n to be 2k+1 and see where it goes from there. you'll probably end up with something = 8m, then you'll see what cases you have.

/edit: Z+ is positive integers right? either you wrote it wrong or it's false. for n = 1, there is no m in Z+ that makes that true. or does Z+ mean non-negative integers? i forget.

This post was edited by carteblanche on Oct 18 2015 10:57am

Let z be a member of Z+ and n odd as given.
We know n=2k+1 for some integer k since n is odd.
Let m be and integer and Thus we have
N^2=8m+1
(2k+1)^2=8m+1
4k^2+4k+1=8m+1
4k^2+4k=8m
K^2+k=2m

From here we can see we have three cases that K=0 which is solved easily to show M=0.
Second case for k=odd gives us odd*odd+odd=even
Or
Third case for k=even gives us even*even+even=even
Thus we can see for any three cases this holds true.