Inequality Proofs

[nycmathguy]

 

[MathLover1]

63.
https://www.math-forums.com/threads/real-numbers-a-b-c.441909/#post-1099802

somewhere must be 62 too, or use 63 to solve 62 in similar way

 

[nycmathguy]

63.
https://www.math-forums.com/threads/real-numbers-a-b-c.441909/#post-1099802

somewhere must be 62 too, or use 63 to solve 62 in similar way

I want you to do 62. Great study notes, you know.

 

[MathLover1]

62.
show that for all real numbers a and b, we have |a| - |b| <= |a-b|

we start with what we know:
|a|+|b| ≥|a−b|

Now, substitute a=a and b=a−b to get,

|a−b|≥|b|−|a| ......................(1)

Also, we can writ |a|+|b|≥|b−a| as |a−b|=|−(a−b)|=|b−a|

Now, substitute a=b−a and b=b to get, |b−a| ≥ |a|−|b|....................(2)

Now, as |b−a|=|a−b|

We may write (2) as |a−b|≥|a|−|b|.............(3)

because |a−b|=|−(a−b)|=|b−a|

From (1) and (3), we can conclude that

|a−b| ≥ ||a|−|b|| or vice verse |a|−|b|| ≤|a−b|

 

[nycmathguy]

62.
show that for all real numbers a and b, we have |a| - |b| <= |a-b|

we start with what we know:
|a|+|b| ≥|a−b|

Now, substitute a=a and b=a−b to get,

|a−b|≥|b|−|a| ......................(1)

Also, we can writ |a|+|b|≥|b−a| as |a−b|=|−(a−b)|=|b−a|

Now, substitute a=b−a and b=b to get, |b−a| ≥ |a|−|b|....................(2)

Now, as |b−a|=|a−b|

We may write (2) as |a−b|≥|a|−|b|.............(3)

because |a−b|=|−(a−b)|=|b−a|

From (1) and (3), we can conclude that

|a−b| ≥ ||a|−|b|| or vice verse |a|−|b|| ≤|a−b|

Very cool. I will study this proof when time allows. My weekend is over just like that. I graduated from high school in 1984. I then took a few steps after graduation and find myself at 57 years old. How on earth did time go by so fast?