This is a compilation of common arithmetic and conceptual mistakes in most lower division math courses. These are conveyed with examples (rather than general forms) and brief explanations for each topic.
Algebra
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Zero factorial does not equal zero
\(0! \neq 0\) because
n factorial equals the fraction with numerator n plus one factorial, and denominator n plus one
\(n! = \frac{(n+1)!}{n+1}\) so
Zero factorial equals the fraction with numerator zero plus one factorial, and denominator zero plus one, which equals one
\(0! = \frac{(0+1)!}{0+1} = 1\)
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One divided by zero does not equal zero
\(1/0 \neq 0\). Observe the graph of
f of x equals one divided by x
\(f(x) = 1/x\) at
x equals zero
\(x=0\).
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x squared times x cubed does not equal x to the sixth power
\(x^2 \cdot x^3 \neq x^6\) because
x squared times x cubed equals the quantity x times x, times the quantity x times x times x times x, which equals x to the fifth power
\(x^2 \cdot x^3 = (x \cdot x)(x \cdot x \cdot x) = x^5\)
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The square root of two plus the square root of three does not equal the square root of five
\(\sqrt{2} + \sqrt{3} \neq \sqrt{5}\) (Sometimes you just cannot 'simplify')
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The quantity x plus y squared does not equal x squared plus y squared
\((x+y)^2 \neq x^2 + y^2\) because
The quantity x plus y squared equals the quantity x plus y times the quantity x plus y, which equals the quantity x plus y times x, plus the quantity x plus y times y, which equals x squared plus two x y plus y squared
\((x+y)^2 = (x+y)(x+y) = (x+y)x + (x+y)y = x^2 + 2xy + y^2\)
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Three times the quantity x plus y squared does not equal the quantity three x plus three y squared
\(3(x+y)^2 \neq (3x+3y)^2\) because
Three times the quantity x plus y squared equals three times the quantity x squared plus two x y plus y squared, which equals three x squared plus six x y plus three y squared
\(3(x+y)^2 = 3(x^2 + 2xy + y^2) = 3x^2 + 6xy + 3y^2\)
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The square root of sixteen does not equal plus or minus four
\(\sqrt{16} \neq \pm 4\) because
The square root of sixteen equals the square root of two to the fourth power, which equals two to the power of four halves, which equals two squared, which equals four
\(\sqrt{16} = \sqrt{2^4} = 2^{4/2} = 2^2 = 4\). However, the solutions to
x squared equals sixteen
\(x^2 = 16\) are
x equals four
\(x = 4\) or
x equals negative four
\(x = -4\).
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Similarly, Negative two squared does not equal four
\(-2^2 \neq 4\) but
The quantity negative two squared equals four
\((-2)^2 = 4\). Parentheses are important!
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The equation Two x squared equals x
\(2x^2 = x\) is not the same as
Two x equals one
\(2x = 1\) because
Two x equals one
\(2x = 1\) does not have the same solutions as
Two x squared equals x
\(2x^2 = x\). This 'reduction' does not simplify the equation
Two x squared equals x
\(2x^2 = x\); it changes it.
Calculus
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The limit as x approaches two of the fraction with numerator x squared minus four, and denominator x minus two, does not equal the fraction with numerator the quantity x minus two times the quantity x plus two, and denominator x minus two
\(\lim_{x \to 2} \frac{x^2-4}{x-2} \neq \frac{(x-2)(x+2)}{x-2}\), but
The limit as x approaches two of the fraction with numerator x squared minus four, and denominator x minus two, equals the limit as x approaches two of the fraction with numerator the quantity x minus two times the quantity x plus two, and denominator x minus two, which equals four
\(\lim_{x \to 2} \frac{x^2-4}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = 4\). Cannot drop the limit between steps or else the statement does not make sense.
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The derivative with respect to x of the natural log of x squared does not equal one divided by x squared
\(\frac{d}{dx} \ln(x^2) \neq \frac{1}{x^2}\) because
The derivative with respect to x of the natural log of x squared equals the derivative with respect to x of two times the natural log of x, which equals two divided by x
\(\frac{d}{dx} \ln(x^2) = \frac{d}{dx} (2\ln(x)) = \frac{2}{x}\)
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Also, The integral of one divided by x with respect to x equals the natural log of x
\(\int \frac{1}{x} dx = \ln(x)\) does not imply
The integral of one divided by x squared with respect to x equals the natural log of x squared
\(\int \frac{1}{x^2} dx = \ln(x^2)\)
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The derivative with respect to x of e to the power of x does not equal x times e to the power of x minus one
\(\frac{d}{dx} e^x \neq x e^{x-1}\) because x is a variable; it is not fixed.
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The derivative with respect to x of e to the power of four x does not equal e to the power of four x
\(\frac{d}{dx} e^{4x} \neq e^{4x}\) but rather
four times e to the power of four x
\(4e^{4x}\)
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The integral of the square root of x with respect to x equals two thirds times x to the power of three halves
\(\int \sqrt{x} dx = \frac{2}{3}x^{3/2}\) does not imply
The integral of the square root of x squared plus one with respect to x equals two thirds times the quantity x squared plus one to the power of three halves
\(\int \sqrt{x^2+1} dx = \frac{2}{3}(x^2+1)^{3/2}\)
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Using L'Hospital's Rule: It states, "Let lim stand for the limit
The limit as x approaches c
\(\lim_{x \to c}\),
The limit as x approaches c from the left
\(\lim_{x \to c^-}\),
The limit as x approaches c from the right
\(\lim_{x \to c^+}\),
The limit as x approaches infinity
\(\lim_{x \to \infty}\), or
The limit as x approaches negative infinity
\(\lim_{x \to -\infty}\), and suppose that
The limit of f of x
\(\lim f(x)\) and
The limit of g of x
\(\lim g(x)\) are both zero or are both positive infinity. If
The limit of the fraction with numerator f prime of x and denominator g prime of x
\(\lim \frac{f'(x)}{g'(x)}\) has a finite value or if the limit is plus or minus infinity, then
The limit of the fraction with numerator f of x and denominator g of x equals the limit of the fraction with numerator f prime of x and denominator g prime of x
\(\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}\)."
This means one cannot simply state
The limit of the fraction with numerator f of x and denominator g of x equals the limit of the fraction with numerator f prime of x and denominator g prime of x
\(\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}\); all the premises must be stated before the conclusion.
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The integral of x, plus two
\(\int x + 2\) is ambiguous; this could be interpreted as
The integral of x with respect to x, plus two
\(\int x dx + 2\).
The integral of the quantity x plus two with respect to x
\(\int (x+2) dx\) is much more clear. And remember "+C" at the end of the solution!
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The integral of sine of theta with respect to x
\(\int \sin(\theta) dx\) does not make sense. The variables must match!
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Finally, "Euler" is pronounced "Oy-ler" not "You-ler".
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