What happens to the range when multiplying inequalities?

[trinvert]

Hi, and thanks in advance for any help.

I'm multiplying two inequalities: [math]g_1(n) ≤ ch_1(n) \text{ for all } n ≥ n_0[/math] and [math]g_2(n) ≤ c'h_2(n) \text{ for all } n ≥ n_0'[/math] where [imath]c, c', n_0, \text{ and } n_0''[/imath] are all constants.
The product of the inequalities is [math]g_1(n)g_2(n) ≤ cc'h_1(n)h_2(n)[/math] but what happens to the range? I'm thinking it's either [math]n^{2} ≥ n_0 n_0'[/math] if you just multiply the ranges, or maybe [math]n ≥ max(n_0, n_0')[/math] if you're taking where the ranges overlap.

Any help and insight into why the answer is what it is would be greatly appreciated!

Edit: It's known that [imath]n ≥ 0[/imath].

 

[lev888]

Hi, and thanks in advance for any help.

I'm multiplying two inequalities: [math]g_1(n) ≤ ch_1(n) \text{ for all } n ≥ n_0[/math] and [math]g_2(n) ≤ c'h_2(n) \text{ for all } n ≥ n_0'[/math] where [imath]c, c', n_0, \text{ and } n_0''[/imath] are all constants.
The product of the inequalities is [math]g_1(n)g_2(n) ≤ cc'h_1(n)h_2(n)[/math] but what happens to the range? I'm thinking it's either [math]n^{2} ≥ n_0 n_0'[/math] if you just multiply the ranges, or maybe [math]n ≥ max(n_0, n_0')[/math] if you're taking where the ranges overlap.

Any help and insight into why the answer is what it is would be greatly appreciated!

Edit: It's known that [imath]n ≥ 0[/imath].

Let's say c = 1, c' = 1, both g functions are -2, both h functions are 1 (constant functions).

According to your reasoning the product of the inequalities would be: 4 ≤ 1.

 

[Dr.Peterson]

Hi, and thanks in advance for any help.

I'm multiplying two inequalities: [math]g_1(n) ≤ ch_1(n) \text{ for all } n ≥ n_0[/math] and [math]g_2(n) ≤ c'h_2(n) \text{ for all } n ≥ n_0'[/math] where [imath]c, c', n_0, \text{ and } n_0''[/imath] are all constants.
The product of the inequalities is [math]g_1(n)g_2(n) ≤ cc'h_1(n)h_2(n)[/math] but what happens to the range? I'm thinking it's either [math]n^{2} ≥ n_0 n_0'[/math] if you just multiply the ranges, or maybe [math]n ≥ max(n_0, n_0')[/math] if you're taking where the ranges overlap.

Any help and insight into why the answer is what it is would be greatly appreciated!

Edit: It's known that [imath]n ≥ 0[/imath].

What do you mean by "the range of an inequality"?

I suspect you are referring to the conditions given for each inequality, which you might call their domains, not ranges.

But you don't multiply domains of functions when you multiply the functions, and you don't multiply conditions when you combine statements.

Rather, you are told that one inequality holds when [imath]n\ge n_0[/imath], and the other holds when [imath]n\ge n'_0[/imath]. So you can only make a conclusion when both conditions are true, which is when [imath]n\ge n_0\wedge n\ge n'_0[/imath], which means [imath]n\ge \max(n_0,n'_0)[/imath] as you said in your second guess.

But also, as lev888 said, you can't multiply inequalities unless you know that all the numbers involved are positive, so your conclusion isn't even correct under this combined condition, unless you omitted some facts about the functions.

When you ask "why the answer is what it is", what answer are you asking about?

 

[trinvert]

What do you mean by "the range of an inequality"?

I suspect you are referring to the conditions given for each inequality, which you might call their domains, not ranges.

But you don't multiply domains of functions when you multiply the functions, and you don't multiply conditions when you combine statements.

Rather, you are told that one inequality holds when [imath]n\ge n_0[/imath], and the other holds when [imath]n\ge n'_0[/imath]. So you can only make a conclusion when both conditions are true, which is when [imath]n\ge n_0\wedge n\ge n'_0[/imath], which means [imath]n\ge \max(n_0,n'_0)[/imath] as you said in your second guess.

But also, as lev888 said, you can't multiply inequalities unless you know that all the numbers involved are positive, so your conclusion isn't even correct under this combined condition, unless you omitted some facts about the functions.

When you ask "why the answer is what it is", what answer are you asking about?

Thanks a lot. Yes, I meant the domains, or conditions. Your explanation makes sense.

Let's say c = 1, c' = 1, both g functions are -2, both h functions are 1 (constant functions).

According to your reasoning the product of the inequalities would be: 4 ≤ 1.

Good point; this is in the context of asymptotic analysis for computer science algorithms, so we can assume everything is positive.

 

[Steven G]

-5 < 2 and -6 < 3. These two inequalities are clearly true.
If I multiply them (whatever that means) I get (-5)(-6) < (2)(3). That is 30 < 6 which is nonsense