Less than or equal sign
If I know for two numbers a and b that
$${a < b }$$
Then is it correct to say that
$$ a \leq b $$
I know that the second statement is true as long as the first one is. It seems OK as it is true but from the other side it seems kinda weird (to me), to say that, if you know, that a is strictly less that b.It's like you lose some extra information.
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$\begingroup$You have a sound understanding of the matter:
It is true that $a\lt b \implies a\leq b$, since $$a \leq b \iff (a\lt b\;\text{or}\;a = b)$$
If you need to prove $a \leq b$, then it suffices to prove either $a < b$ or $a = b$.
On the other hand, if you need (or want) to prove $a$ is strictly less than $b$, then it is necessary to prove $a < b$.
There can be, indeed, some loss of information, just as we lose information when, while knowing $x = 3$, we assert only the (true) claim that $x \leq 3$.
But there are examples where we can make a claim for a wider range of values by using the inclusive $\leq$.
I don't know if this example will be of any help, but if you are trying to prove, by induction, that $n^2 \leq 2^n$ for $n>3$, in your inductive proof, we first have the base case $n = 4$, where the inequality holds precisely because equality holds: $4^2 = 2^4$, hence $4^2 \leq 2^4$ is true.
But in the inductive step, you might use the following chain: $$(n+1)^2 = n^2 + 2n + 1 \leq 2^n + \underbrace{2n + 1}_{\large <\; 2^n,\,n\geq 3} \lt 2^n + 2^n = 2\cdot 2^n = 2^{n+1}$$
So in the inductive step, even though we find we have a strict inequality involved, the point of the induction is to prove $n^2 \leq 2^n$, for $n \gt 3$ and because the base case is true (because equality holds), we can actually say more by affirming the NON-strict inequality (since the range of $n$ for which the proposition is true is greater with $\leq$ than with $\lt$.)
In the end, the amount of information conveyed by the choice of using $\lt$ vs. $\leq$ depends, as do many choices, on context.
$\endgroup$ 0 $\begingroup$It is correct to say that if $a < b$, then $a \le b$. And yes, I completely agree with you: if explicitly being shown $a < b$, there is a certain "loss" of information when we write $a \le b$ instead of being concrete.
"$\le$" is, however, very useful in real-life, practical mathematical problems such as optimization when you want to say that one quantity does not exceed another.
$\endgroup$ $\begingroup$Yes. Since $a\leq b \Leftrightarrow (a<b \vee a=b)$, $a<b$ implies $a\leq b$.
$\endgroup$ $\begingroup$Think about it as "can be". For instance:
a≤b means: can be that a<b, or can be that a=b, or can be that a
Example 1: Say x is any real number so that x≤3
Then x can be less than 3, or x can be 3, or x can be anywhere less than 3 up to 3.
In this example, any option will satisfy the initial statement.
Example 2: Say 2<3, then 2≤3
2≤3 means: it can be that 2<3, or can be 2=3, or 2<3 and 2=3.
Since we found that 2≤3 can be 2<3 as true, 2<3 implies 2≤3.
Example 3: (if it is still confusing) I want lemon juice
You go to the kitchen and find orange juice and lemon juice. What do you give me? You can give me orange juice, or lemon juice, or both.
Since I only want lemon juice, your only option is lemon juice. So even though this example is silly, it shows that what you see is more than what they ask. Using OR means that it "can be" as an option. You always have to count for the inclusive part. For example, the probability of A or B, you need to subtract the probability of A and B if you want independent probabilities.
I know it is a lot of examples but I hope it helps.
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