MHB Math Trick: How to Turn 1/25 Into 10 - Step by Step Guide

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The discussion clarifies that 1/25 does not directly become 10; instead, the function f(x) = 2/√x evaluates to 10 when x is 1/25. By calculating f(1/25), it is shown that √(1/25) equals 1/5, leading to f(1/25) = 2/(1/5), which simplifies to 10. Participants express understanding of the mathematical steps involved. The explanation emphasizes the importance of function evaluation rather than direct transformation. This highlights a common misconception in interpreting mathematical functions.
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How it possible for 1/25 to become 10? May anyone show me the steps please
 

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Hi Teh,

It's not that $1/25$ became $10$, but that $f(1/25) = 10$. For since $f(x) = 2/\sqrt{x}$ and $\sqrt{1/25} = 1/5$, then $f(1/25) = 2/(1/5) = 2 \cdot 5 = 10$.
 
Teh said:
How it possible for 1/25 to become 10? May anyone show me the steps please

$\frac{1}{25}$ does not become 10

but $f(\frac{1}{25}) = \frac{2}{\sqrt{\frac{1}{25}}} = \frac{2}{\frac{1}{5}} = 2 * 5 =10 $
 
ahhh okay plug it into the square root of x thank you very much :D
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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