# Proof Inequality: x ≤ y | Homework Statement

• lynchu
In summary, the conversation discusses a proof involving real numbers x and y, where if x is less than or equal to y plus any positive real number, then x is also less than or equal to y. However, there is a discrepancy regarding whether x is strictly less than y or less than or equal to y. The conversation goes on to discuss the need to show a contradiction in order to prove the statement.
lynchu

## Homework Statement

Let x and y be real numbers. Prove that if x =< y + k for every positive real number k, then x =< y

## The Attempt at a Solution

x =< y + k
-y + x =< k
since k is positive, the lowest value it can take doesn't include 0: -y + x < 0
x < y

So I get x < y from x =< y + k and not the required x =< y. Am I right or I'm screwing up somewhere? Thanks for your help.

If x and y are equal, then $x \leq y + k$ for all $k>0 \in \mathbb R [/tex], right? So your conclusion must not be right. Try this. Assume there are an x and y such that [itex] x \leq y + k$ for all $k>0$ but $x > y$ and show this leads to a contradiction.

Out of x > y I got -y + x > 0.
So that 0 < -y + x =< k

But that's not quite a contradiction? Or is it?

I haven't studied this at all, but intuitively I would say that lynchu is correct.

hgfalling said:
If x and y are equal, then $x \leq y + k$ for all $k>0 \in \mathbb R [/tex], right? If x and y are equal, then we only have x=y+k in the case that k=0, now if k>0 then x<y+k. Right? Mentallic said: I haven't studied this at all, but intuitively I would say that lynchu is correct. If x and y are equal, then we only have x=y+k in the case that k=0, now if k>0 then x<y+k. Right? Well yes, but that's not the statement that's being evaluated. The theorem to be proved is: IF [itex] x \leq y + k$ for all real $k>0$, THEN $x \leq y$.

lynchu's claim was stronger than this; he claimed that in fact

IF $x \leq y + k$ for all real $k>0$, THEN $x < y$.

But this isn't true, because if x and y are equal, then the condition holds, but the conclusion is false.

What you said was:

IF $x=y$, THEN $x<y+k$ for all real $k>0$.

lynchu said:
Out of x > y I got -y + x > 0.
So that 0 < -y + x =< k

But that's not quite a contradiction? Or is it?

No, not really. All you need to do is show me ONE k value so that if $x \leq y + k$ and $x > y$ you get a contradiction.

Oh I see, yeah now it makes sense why it should be $$x\leq y$$

Appreciate the help so far.
I've spent way too much time on this and yet I just don't see it.

## 1. What is the definition of proof inequality?

Proof inequality is a mathematical concept that compares two numbers or expressions using the symbols <, >, ≤, or ≥. It states that one number or expression is less than, greater than, less than or equal to, or greater than or equal to the other.

## 2. How do you prove an inequality?

To prove an inequality, you must show that the statement is true for all possible values of the variables involved. This can be done by using algebraic manipulation, logical reasoning, or mathematical theorems.

## 3. What are the rules for solving inequalities?

The rules for solving inequalities are similar to those for solving equations. However, when multiplying or dividing by a negative number, the direction of the inequality sign must be reversed. Additionally, when taking the square root of both sides, the positive and negative solutions must be considered.

## 4. How do you determine the solution set for an inequality?

The solution set for an inequality is the set of all values that satisfy the inequality. This can be found by graphing the inequality on a number line or by plugging in values to determine which ones make the statement true.

## 5. What are some real-life applications of proof inequality?

Proof inequality has numerous real-life applications, such as in economics to compare prices, in physics to analyze relationships between variables, and in statistics to compare data sets. It is also commonly used in problem-solving to determine the best course of action based on given constraints.

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