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29 Ιουλ 2010 · The equality $nx = x^2$ is not comparable to an equality like $\sin x^2 = 1 - \cos x^2$ : the equation $nx = x^2$ holds for only two values of $x$, where as the trigonometric equation holds for any real/complex value of $x$.
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23 Οκτ 2015 · I need to find the flaw in the following proof: $a,b\in\mathbb{R}$\ $\left\{ 0 \right\} $ such that $a=b$ 1) Multiplying both sides by $a$ yields the equality: $a^2=ab$ 2) Subtracting $b^2$ from...
Error. Usually, if a proof proves a statement that is clearly false, the proof has probably divided by zero in some way. In this case, the quantity of is as , since one cannot divide by zero, the proof is incorrect from that point on. Thus, this proof is false.
We can find the first false inference by finding the first false equation above; if it is equation number n + 1, then the inference from equation n to n + 1 must be incorrect. Doing this above we infer that the culprit is " via cancel 3 − 3 ", i.e. x ⋅0 = y ⋅0 ⇏ x = y, i.e. it divided by 0.
1=2: A Proof using Beginning Algebra. The Fallacious Proof: Step 1: Let a = b . Step 2: Then , Step 3: , Step 4: , Step 5: , Step 6: and . Step 7: This can be written as , Step 8: and cancelling the from both sides gives 1=2. See if you can figure out in which step the fallacy lies.
You can easily find you have a fallacy in your statement if you idenitfy the following results in your math-script. Your proof is being theoretically correct, and no mistakes are found. Your proof ended with some equals, which are universally unequal. For example, 2 = 1, a = b, where a > b etc..,.
The Fallacious Proof: Step 1: -1/1 = 1/-1. Step 2: Taking the square root of both sides: sqrt (-1/1) = sqrt (1/-1) (where "sqrt" denotes the square-root symbol which cannot be displayed on text-only browsers.) Step 3: Simplifying: sqrt (-1) / sqrt (1) = sqrt (1) / sqrt (-1) Step 4: In other words, i/1 = 1/i. Step 5: Therefore, i / 2 = 1 / (2i),
