Why is 7 less than or equal to 7 in inequality math?

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In inequality math, the expression 7 ≤ 7 indicates that 7 is either less than or equal to 7. The "or" in this context allows for both possibilities to be considered true. While 7 < 7 is false, the statement 7 = 7 is true, satisfying the condition for the inequality. Therefore, 7 ≤ 7 is valid because at least one part of the inequality holds true. This illustrates the fundamental principle that inequalities can include equalities.
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Basically I was reviewing a proof of one of the inequality properties and there was a statement that a <= a , or in other words for ex. 7<=7. So my question is why is that, since 7 is really = to 7, at least I think so.

thanks.
 
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The sign \leq mean less than or equal to. So saying 7 \leq 7 means that 7 is less than or equal to 7. Now, in mathematics (and pretty much everywhere else) the word "or" means that either one or the other holds (and depending on the context, potentially both). Clearly 7 &lt; 7 is absurd, but 7 = 7 is true which means that one of the two clauses holds, thus 7 \leq 7.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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