Unit P4: Pure Mathematics A2 4

A proof by contradiction begins by assuming the exact negation of the statement to be proved. If valid deductions from that assumption force an impossibility or conflict with a given condition, the assumption is false and the original statement must be true.

Use four explicit moves: (1) state the contradictory assumption; (2) translate it into algebra, parity, factors or an inequality; (3) derive a named contradiction; (4) reject the assumption and state the original conclusion. For an implication ‘if P, then Q’, assume P is true and Q is false—not merely that P is false.

For $\sqrt2$, assume $\sqrt2=a/b$ where integers $a,b$ have no common factor. Then $a^2=2b^2$, so $a$ is even; write $a=2k$. Now $4k^2=2b^2$, hence $b^2=2k^2$ and $b$ is even. Thus $a$ and $b$ share factor 2, contradicting lowest terms. Therefore $\sqrt2$ is irrational.

For infinitely many primes, assume the complete list is $p_1,\ldots,p_n$ and form $N$ above. Dividing $N$ by any listed prime leaves remainder 1. Yet $N>1$ has a prime divisor, which is therefore absent from the supposedly complete list. This contradiction proves that there are infinitely many primes.

Partial fractions rewrite one rational function as a sum of simpler rational terms. The denominator determines the complete template; once the coefficients are found, the equivalent form is easier to integrate, differentiate or expand as a series.

Factor the denominator, divide first if the fraction is improper, and write every required term. Multiply through by the full denominator to form a polynomial identity. Substitute roots of the linear factors to isolate coefficients, then compare coefficients or use another convenient value for any coefficient left. Recombine the result over the original denominator to check it.

For example, $$\frac{5x+1}{(x-1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}.$$ Multiplying through gives $$5x+1=A(x+2)^2+B(x-1)(x+2)+C(x-1).$$ Setting $x=1$ gives $A=2/3$; setting $x=-2$ gives $C=3$; comparing $x^2$ coefficients gives $B=-2/3$.

Parametric equations give both coordinates in terms of a third variable: $x=f(t)$ and $y=g(t)$. Each permitted value of $t$ produces a point $(x,y)$; changing $t$ traces the curve. A Cartesian equation removes $t$ and links $x$ directly to $y$.

To convert to Cartesian form, make $t$ or a simple function of $t$ the subject of one equation and substitute into the other. For trigonometric parameters, isolate $\sin t$ and $\cos t$ and use $\sin^2t+\cos^2t=1$. Finally translate the stated range of $t$ into restrictions on $x$ or $y$.

For $x=(t+3)/(t+1)$ and $y=2/(t+1)$ with $t>0$, notice that $x=1+2/(t+1)=1+y$. Hence $y=x-1$. Since $t+1>1$, $0<y<2$ and therefore $1<x<3$. The unrestricted line would contain points that the parameter never reaches.

For $x=3+2\cos t$ and $y=-1+4\sin t$, $$\cos t=\frac{x-3}{2},\qquad \sin t=\frac{y+1}{4}.$$ Therefore $$\frac{(x-3)^2}{4}+\frac{(y+1)^2}{16}=1.$$ The allowed interval for $t$ decides whether this represents the whole ellipse or only part of it.

The binomial series expands a rational power around $x=0$. Put the expression into the standard form $(1+u)^n$ first; the coefficients come from the descending products of $n$, while the size of $u$ controls where a non-terminating series is valid.

Rewrite $(ax+b)^n=b^n\left(1+\frac{a}{b}x\right)^n$. For a negative or non-integer rational $n$, require $\left|\frac{a}{b}x\right|<1$, so $|x|<\left|b/a\right|$. A non-negative integer power terminates and does not need this convergence restriction.

For $(4+5x)^{1/2}$, take out $4^{1/2}=2$ and use $u=5x/4$: $$2\left(1+\frac12u-\frac18u^2+\frac1{16}u^3+\cdots\right)=2+\frac54x-\frac{25}{64}x^2+\frac{125}{512}x^3+\cdots.$$ This series is valid for $|5x/4|<1$, or $|x|<4/5$.

For a rational function, decompose first and expand each simple fraction. For example, $$\frac{2}{1-2x}+\frac{3}{4+x}=2(1-2x)^{-1}+\frac34(1+x/4)^{-1}$$ $$=\frac{11}{4}+\frac{61}{16}x+\frac{515}{64}x^2+\cdots.$$ The two component ranges are $|x|<1/2$ and $|x|<4$, so the combined expansion requires the narrower condition $|x|<1/2$.

A curve need not be written as $y=f(x)$ before its gradient can be found. For an implicit equation, differentiate every term with respect to $x$ and attach $dy/dx$ whenever a function of $y$ is differentiated. For a parametric curve, compare how both coordinates change with the parameter.

For $x^2+xy+y^2=7$, differentiation gives $$2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0,$$ so $$\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$$ At $(1,2)$ the tangent gradient is $-4/5$; the normal gradient is $5/4$.

For $x=t^2+1$ and $y=t^3-3t$, $$\frac{dy}{dx}=\frac{3t^2-3}{2t}.$$ At $t=2$, the point is $(5,2)$ and the gradient is $9/4$, so the tangent is $$y-2=\frac94(x-5).$$

At a point $(x_0,y_0)$, use $y-y_0=m(x-x_0)$. A finite non-zero tangent gradient $m$ gives normal gradient $-1/m$. Keep the point and gradient exact until the final line is simplified.

A differential equation states how a quantity changes, rather than giving the quantity directly. Build it by naming the changing variables, translating each rate statement with its sign and units, and differentiating the model relation with respect to the stated independent variable.

Suppose a tank has constant horizontal area $10\,\mathrm{m^2}$, water enters at $5\,\mathrm{m^3\,min^{-1}}$, and leaves at $0.2h\,\mathrm{m^3\,min^{-1}}$ when the depth is $h$ metres. Since $V=10h$, $$\frac{dV}{dt}=5-0.2h\quad\text{and}\quad\frac{dV}{dt}=10\frac{dh}{dt},$$ so the model is $10\,dh/dt=5-0.2h$.

For connected rates: (1) express all geometry in one changing variable; (2) differentiate before inserting the instant's values; (3) connect rates with the chain rule; (4) solve for the requested derivative; (5) report its sign and units.

For a cube of side $x$ cm, $S=6x^2$ and $V=x^3$. If its surface area decreases at $12\,\mathrm{cm^2\,s^{-1}}$, then $$-12=12x\frac{dx}{dt}.$$ At $x=3$, $dx/dt=-1/3\,\mathrm{cm\,s^{-1}}$, and $$\frac{dV}{dt}=3x^2\frac{dx}{dt}=-9\,\mathrm{cm^3\,s^{-1}}.$$

Rotating the region between a non-negative curve $y=f(x)$ and the $x$-axis through a full turn produces circular cross-sections. A slice of thickness $dx$ has area $\pi y^2$, so its volume is accumulated by integration.

For $y=x+1$ on $0\le x\le2$, $$V=\pi\int_0^2(x+1)^2\,dx=\pi\left[\frac{(x+1)^3}{3}\right]_0^2=\frac{26\pi}{3}.$$ Square the whole expression for $y$ before integrating, and obtain the limits from the region's $x$-coordinates.

If $x=x(t)$ and $y=y(t)$, replace $dx$ by $x'(t)dt$: $$V=\pi\int_{t_1}^{t_2}y(t)^2x'(t)\,dt.$$ For $x=t^2$, $y=t+1$, $0\le t\le2$, $$V=\pi\int_0^2(t+1)^2(2t)\,dt=\frac{68\pi}{3}.$$ Choose the parameter limits in the direction of increasing $x$; if $x'(t)<0$ throughout, reverse the limits so the volume is positive.

Integration by substitution reverses the chain rule; integration by parts reverses the product rule. First identify which structure is present, then transform every factor, differential and—when the integral is definite—limit consistently.

For $$\int x\sqrt{x^2+4}\,dx,$$ let $u=x^2+4$, so $du=2x\,dx$. Then $$\int x\sqrt{x^2+4}\,dx=\frac12\int u^{1/2}\,du=\frac13(x^2+4)^{3/2}+C.$$ For a definite integral, either change both limits to $u$-values or return to $x$ before applying the original limits—never mix the two.

For $\int x^2e^x\,dx$, take $u=x^2$ and $dv=e^x dx$. Two applications give $$\int x^2e^x\,dx=x^2e^x-2\int xe^x\,dx=e^x(x^2-2x+2)+C.$$ Also, writing $\ln x$ as $1\cdot\ln x$ gives the required result $$\int\ln x\,dx=x\ln x-x+C.$$

When a proper rational expression has factorised linear denominators, decompose it into simpler fractions before integrating. Match a constant numerator to each linear factor and an additional term for every repeated power.

For $$\frac{5x+1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2},$$ matching numerators gives $A=-4$ and $B=9$. Hence $$\int\frac{5x+1}{(x+1)(x+2)}\,dx=-4\ln|x+1|+9\ln|x+2|+C.$$ Differentiate the result to check both coefficients.

A repeated linear factor does not produce another logarithm: $$\int\frac{3}{(x-1)^2}\,dx=-\frac{3}{x-1}+C.$$ Likewise, $$\int\frac{2}{(2x-1)^4}\,dx=-\frac{1}{3(2x-1)^3}+C.$$

A first-order equation is separable when all $y$-dependence can be placed with $dy$ and all $x$-dependence with $dx$. Integrating produces a family of curves—the general solution. An initial or boundary condition then selects a particular member.

For $dy/dx=2xy$ with $y\ne0$, separation gives $$\frac1y\,dy=2x\,dx,$$ so $$\ln|y|=x^2+C\quad\Longrightarrow\quad y=Ae^{x^2}.$$ This non-zero family, together with the equilibrium solution $y=0$ lost when dividing by $y$, is the general solution.

If $y(0)=3$, then $3=Ae^0$, so the particular solution is $$y=3e^{x^2}.$$ Check it by differentiating: $dy/dx=6xe^{x^2}=2x(3e^{x^2})=2xy$, and it satisfies the initial condition.

A reliable sequence is: separate with $dy$ and $dx$ visible; integrate both sides and include one arbitrary constant; simplify only when it helps; use the condition; state the requested form; differentiate to verify. Partial fractions or integration by parts may be needed after separation.

For a parametric curve $x=x(t)$, $y=y(t)$, convert the Cartesian area element $y\,dx$ using $dx=x'(t)dt$. The parameter values at the region's endpoints supply the limits; the curve does not need to be rearranged into $y=f(x)$.

For $x=t^2$, $y=t+1$, $0\le t\le2$, $x'(t)=2t$ and the curve lies above the $x$-axis. Thus $$A=\int_0^2(t+1)(2t)\,dt=\left[\frac23t^3+t^2\right]_0^2=\frac{28}{3}.$$

A vector records both magnitude and direction. In two dimensions it has two components; in three dimensions it has three. The components measure the signed change parallel to each coordinate axis.

Two vectors are equal when their corresponding components are equal, even if they start at different points. Thus $\begin{pmatrix}3\\-2\\5\end{pmatrix}$ and $3\mathbf i-2\mathbf j+5\mathbf k$ represent the same vector.

The magnitude $|\mathbf a|$ is the length of a vector. It follows from Pythagoras in two or three dimensions. Dividing a non-zero vector by its magnitude keeps its direction but changes its length to $1$.

For $\mathbf a=\begin{pmatrix}4\\-4\\2\end{pmatrix}$, $$|\mathbf a|=\sqrt{4^2+(-4)^2+2^2}=6.$$ A unit vector in the same direction is $$\widehat{\mathbf a}=\frac16\begin{pmatrix}4\\-4\\2\end{pmatrix}=\begin{pmatrix}2/3\\-2/3\\1/3\end{pmatrix}.$$ Its magnitude is $1$, which checks the calculation.

Vector addition combines successive displacements, while scalar multiplication changes a vector's length and possibly its direction. Both operations are performed component by component.

If $\mathbf a=\begin{pmatrix}2\\-1\\3\end{pmatrix}$ and $\mathbf b=\begin{pmatrix}-4\\5\\1\end{pmatrix}$, then $$\mathbf a+\mathbf b=\begin{pmatrix}-2\\4\\4\end{pmatrix},\qquad 2\mathbf a-\mathbf b=\begin{pmatrix}8\\-7\\5\end{pmatrix}.$$ Geometrically, addition places vectors head-to-tail; the resultant joins the starting point to the finishing point.

The position vector of a point runs from the fixed origin to that point. If $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$, the displacement from $A$ to $B$ is found by subtracting start from finish.

If $A=(2,-3,5)$ and $B=(8,1,-1)$, then $$\overrightarrow{AB}=\begin{pmatrix}8\\1\\-1\end{pmatrix}-\begin{pmatrix}2\\-3\\5\end{pmatrix}=\begin{pmatrix}6\\4\\-6\end{pmatrix}.$$ Reversing the journey gives $\overrightarrow{BA}=-\overrightarrow{AB}$.

Position vectors also encode division of a segment. If $B$ lies one third of the way from $A$ to $C$, then $$\mathbf b=\mathbf a+\frac13(\mathbf c-\mathbf a),$$ so $\mathbf c=3\mathbf b-2\mathbf a$. The coefficients reflect the stated direction and ratio.

The distance between two points is the magnitude of the displacement vector joining them. Subtract corresponding coordinates, square the changes, add them, and take the non-negative square root.

For $P=(1,-2,4)$ and $Q=(5,1,-8)$, $$\overrightarrow{PQ}=\begin{pmatrix}4\\3\\-12\end{pmatrix},$$ so $$PQ=\sqrt{4^2+3^2+(-12)^2}=\sqrt{169}=13.$$ The same distance results from $\overrightarrow{QP}$ because every component is squared.

A vector line is determined by one point and one non-zero direction vector. The parameter moves through every real value, generating every point on the line.

Through $C=(1,2,0)$ and $D=(3,-1,4)$, a direction is $\overrightarrow{CD}=(2,-3,4)$, so $$\mathbf r=\begin{pmatrix}1\\2\\0\end{pmatrix}+t\begin{pmatrix}2\\-3\\4\end{pmatrix}.$$ At $t=1$ this gives $D$, providing a quick check.

The scalar product converts two vectors into a number that measures directional alignment. Its component form calculates the value; its geometric form connects that value to the angle between the vectors.

For non-zero $\mathbf a=\begin{pmatrix}1\\2\\2\end{pmatrix}$ and $\mathbf b=\begin{pmatrix}2\\0\\1\end{pmatrix}$, $$\mathbf a\cdot\mathbf b=4,\quad |\mathbf a|=3,\quad|\mathbf b|=\sqrt5,$$ so $$\theta=\cos^{-1}\!\left(\frac{4}{3\sqrt5}\right)\approx53.4^\circ.$$ For the acute angle between two lines, use the absolute value of the cosine because either direction vector may be reversed.

If non-zero vectors satisfy $\mathbf a\cdot\mathbf b=0$, then $\cos\theta=0$ and they are perpendicular. For example, $(1,2,2)\cdot(2,1,-2)=2+2-4=0$. This condition can locate the foot of a perpendicular by making a displacement vector dot a line's direction equal to zero.