Simple Math Theory
- Thread starter Lynxtene
- Start date
ack... don't ever try to just plug in some numbers unless it's for a quick a dirty look at things.
Doing this is bound to get you wrong answers eventually.
The way to look at this is two basic facts
\(\displaystyle \text{$a > c$ and $b > d \Rightarrow a+b > c+d$}\\
\text{$a > c$ and $b > d \Rightarrow a b > c d$}\)
so let's look at the first two statements, let's look at #2 first
\(\displaystyle \text{$m + p > n + q$}\\
\text{Here we have $m > n$ and $p > q$ as given, so we can say yes}\\
m+p > n+q
\)
Now look at #1
\(\displaystyle m+q > n+p\\
\text{Here we have $m>n$ but we do not have $q>p$, in fact we have the opposite}\\
\text{So we cannot say with certainty that $m+q > n+p$}
\)
Apply these same ideas to #3 and #4
Doing this is bound to get you wrong answers eventually.
The way to look at this is two basic facts
\(\displaystyle \text{$a > c$ and $b > d \Rightarrow a+b > c+d$}\\
\text{$a > c$ and $b > d \Rightarrow a b > c d$}\)
so let's look at the first two statements, let's look at #2 first
\(\displaystyle \text{$m + p > n + q$}\\
\text{Here we have $m > n$ and $p > q$ as given, so we can say yes}\\
m+p > n+q
\)
Now look at #1
\(\displaystyle m+q > n+p\\
\text{Here we have $m>n$ but we do not have $q>p$, in fact we have the opposite}\\
\text{So we cannot say with certainty that $m+q > n+p$}
\)
Apply these same ideas to #3 and #4
Dr.Peterson
Elite Member
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The key is that you didn't try negative numbers. You would have been right if the numbers were assumed to be positive.
Trying specific numbers is the right way to find a counterexample (to show it is not always true). It will not prove something is always true, but even there it can help you think about the problem (by not focusing on what the results are, but why). You can also think about separate cases, which is related to trying different numbers.
Trying specific numbers is the right way to find a counterexample (to show it is not always true). It will not prove something is always true, but even there it can help you think about the problem (by not focusing on what the results are, but why). You can also think about separate cases, which is related to trying different numbers.
I led you astray by not considering negative numbers when multiplying. What I wrote certainly is true if all the numbers are positive.
for #3 you could have m > 0, and n, p, q < 0
Then mn < 0, but pq > 0 and thus mn < pq
for #4 I think we have to look at some cases
If all 4 are positive, what I wrote applies and mp > nq with certainty.
If only q is negative then mp>0, nq < 0, and again mp > nq with certainty
If p and q are negative then we can write
m|p| > n|q|
m(-p) > n(-q), thus
mp < nq
and this contradicts the statement we're trying to show. So no certainty.
As an example for this consider (m,n,p,q) = (5,2,-2,-4)
mp = -10, nq=-8
mp < nq
My apologies for not considering this in the original post.
I still maintain your teacher's advice is sketchy to say the least.
Logic > plugging in numbers and seeing what happens.
Last edited:
Steven G
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- Dec 30, 2014
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- 14,561
You beat me to it. How did you (of all people!) not consider negative numbers?
I hope that you have a nice corner to sit in for a negative amount of time.
I am guessing that the OP has lost interest, but some other student may be interested.
When we are asked whether it is true that a specific proposition is always true, we can always say that it is false that that specific proposition is always true by showing a single example where it is not true. That is a very simple kind of proof, but it does not tell you how to find a counter-example. It works if you "see" one.
There is an alternative way that does not involve intuition but does involve some backward thinking. Let's take A as an example.
We want to prove or disprove that [MATH]m > n > p > q \implies (m + q) > (n + p).[/MATH]
We assume a contradiction.
[MATH]\text {ASSUME } m > n > p > q \text { and } (m + q) = (n + p).[/MATH]
[MATH]m > n \implies \exists \ a > 0 \text { such that } m = n + a.[/MATH]
[MATH]n > p \implies \exists \ b > 0 \text { such that } n = p + b.[/MATH]
Now I choose arbitrary m, a, and b except I insist that a and b each be positive.
m = 100, a = 1, and b = 2.
[MATH]m = 100 \text { and } a = 1 \implies 100 = n + 1 \implies n = 99.[/MATH]
[MATH]n = 99 \text { and } b = 2 \implies 99 = p + 2 \implies p = 97.[/MATH]
[MATH]\therefore n + p = 99 + 97 = 196.[/MATH]
But wait! There were two restrictions implied for q, namely m + q = n + p and p > q.
[MATH]m + q = n + p \implies 100 + q = 196 \implies q = 96 < p = 97.[/MATH]
So I just found my counter-example.
[MATH]m = 100,\ n = 99, \ p = 97, \text { and } q = 96 \implies m > n > p > q \ \to \ \text m + q \not > n + p.[/MATH]
[MATH]\therefore m > n > p > q \not \implies (m + q) > (n + p).[/MATH]
The counter-example process does not depend on guessing.