knowLittle said:Homework Statement
Prove for any ##a,b \in \mathbb{R^+} : |a-b| \leq \sqrt{a^2 +b^2} ##
The Attempt at a Solution
How should I start? Can I use induction?
Should I use contrapositive?
knowLittle said:uh, are you serious?
## (|a-b|)^2 = (a^2 - 2ab + b^2 ) \leq (\sqrt{a^2 +b^2})^2 = a^2 +b^2##
Then the LHS will always be lesser for any ## \mathbb{R^+}##
knowLittle said:Wouldn't I need this first:
## -\sqrt{ a^2 + b^2} < (a-b) < \sqrt{ a^2 +b^2 }\\ ##
Then,
## a^2 +b^2 < ( a-b )^2 = (a^2 - 2ab + b^2 ) < (\sqrt{a^2 +b^2})^2 = a^2 +b^2##
Now, I don't understand
I see what you are doing.Curious3141 said:I'm sorry, I do not understand what you're doing here.
The first step is to consider the square of the expression ##|a-b|##.
It should be trivial to recognise that ##(|a-b|)^2 = |(a-b)^2| = a^2 + b^2 - 2ab##
The next step is to note that ##a^2 + b^2 - 2ab < a^2 + b^2## for positive reals a and b.
That gives you ##(|a-b|)^2 < a^2 + b^2##. The final step is to take the square root of both sides.
Can you finish it off from here?
knowLittle said:I see what you are doing.
About what I wrote prior to this...
I was talking about this supposedly property of absolute value
## |a-b| < c ##, then
## -c< a-b <c ##
Induction is a mathematical proof technique that involves proving a statement for a base case (usually n = 1) and then showing that if the statement holds for an arbitrary value of n, it also holds for n+1. By proving the statement for the base case and showing that it holds for n+1, it is implied that the statement holds for all natural numbers. This method can be used to prove the inequality |a-b| ≤ $\sqrt{a^2 + b^2}$ by setting the base case as n = 1 and then showing that if the statement holds for n, it also holds for n+1.
The absolute value in the inequality |a-b| ≤ $\sqrt{a^2 + b^2}$ ensures that the result is always positive. This is important because the square root of a squared term can result in a negative or positive value, but the absolute value ensures that the result is always positive. This is necessary for proving the inequality using induction, as the proof relies on showing that the result holds for all natural numbers.
Yes, the inequality |a-b| ≤ $\sqrt{a^2 + b^2}$ can also be proven using other techniques, such as direct proof or proof by contradiction. However, induction is a commonly used method for proving mathematical inequalities and can be particularly useful when dealing with statements involving natural numbers.
The intuition behind using induction to prove the inequality |a-b| ≤ $\sqrt{a^2 + b^2}$ is that by showing the statement holds for a base case and then showing that if it holds for an arbitrary value, it also holds for the next value, it can be concluded that the statement holds for all natural numbers. This is because the next value can always be reached by adding 1 to the previous value, and if the statement holds for n, it also holds for n+1.
Yes, there are some limitations to using induction to prove the inequality |a-b| ≤ $\sqrt{a^2 + b^2}$. For example, induction can only be used to prove statements that involve natural numbers. Additionally, the base case and the induction step must be carefully chosen in order to prove the statement correctly. Induction can also be a time-consuming process and may not always be the most efficient method for proving an inequality.