Section 2
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( a + b ) + ( c + d ) = ( a + d ) + ( b + c ) {\displaystyle (a+b)+(c+d)=(a+d)+(b+c)\,}
a + b + c + d = ( a + d ) + ( b + c ) {\displaystyle a+b+c+d=(a+d)+(b+c)\,} by associativity
a + b + d + c = ( a + d ) + ( b + c ) {\displaystyle a+b+d+c=(a+d)+(b+c)\,} by commutativity
a + d + b + c = ( a + d ) + ( b + c ) {\displaystyle a+d+b+c=(a+d)+(b+c)\,} by commutativity
( a + d ) + ( b + c ) = ( a + d ) + ( b + c ) {\displaystyle (a+d)+(b+c)=(a+d)+(b+c)\,} by associativity
1 (alternative)
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( a + b ) + ( c + d ) = ( a + d ) + ( b + c ) {\displaystyle (a+b)+(c+d)=(a+d)+(b+c)\,}
= a + ( b + ( c + d ) ) {\displaystyle =a+(b+(c+d))\,} by associativity
= a + ( b + ( d + c ) ) {\displaystyle =a+(b+(d+c))\,} by commutativity
= a + ( d + ( b + c ) ) {\displaystyle =a+(d+(b+c))\,} by associativity
= a + ( d + ( b + c ) ) {\displaystyle =a+(d+(b+c))\,} by associativity
( a + b ) + ( c + d ) = ( a + c ) + ( b + d ) {\displaystyle (a+b)+(c+d)=(a+c)+(b+d)\,}
a + b + c + d = ( a + c ) + ( b + d ) {\displaystyle a+b+c+d=(a+c)+(b+d)\,} by associativity
a + c + b + d = ( a + c ) + ( b + d ) {\displaystyle a+c+b+d=(a+c)+(b+d)\,} by commutativity
( a + c ) + ( b + d ) = ( a + c ) + ( b + d ) {\displaystyle (a+c)+(b+d)=(a+c)+(b+d)\,} by associativity
2 (alternative)
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( a + b ) + ( c + d ) = ( a + c ) + ( b + d ) {\displaystyle (a+b)+(c+d)=(a+c)+(b+d)\,}
a + ( b + ( c + d ) ) = ( a + c ) + ( b + d ) {\displaystyle a+(b+(c+d))=(a+c)+(b+d)\,} by associativity
a + ( ( b + c ) + d ) = ( a + c ) + ( b + d ) {\displaystyle a+((b+c)+d)=(a+c)+(b+d)\,} by associativity
a + ( ( c + b ) + d ) = ( a + c ) + ( b + d ) {\displaystyle a+((c+b)+d)=(a+c)+(b+d)\,}
by commutativity
a + ( c + ( b + d ) ) = ( a + c ) + ( b + d ) {\displaystyle a+(c+(b+d))=(a+c)+(b+d)\,}
by associativity
( a + c ) + ( b + d ) = ( a + c ) + ( b + d ) {\displaystyle (a+c)+(b+d)=(a+c)+(b+d)\,} by associativity
( a − b ) + ( c − d ) = ( a + c ) + ( − b − d ) {\displaystyle (a-b)+(c-d)=(a+c)+(-b-d)\,}
a − b + c − d = ( a + c ) + ( − b − d ) {\displaystyle a-b+c-d=(a+c)+(-b-d)\,} by associativity
a + c − b − d = ( a + c ) + ( − b − d ) {\displaystyle a+c-b-d=(a+c)+(-b-d)\,} by commutativity
( a + c ) + ( − b − d ) = ( a + c ) + ( − b − d ) {\displaystyle (a+c)+(-b-d)=(a+c)+(-b-d)\,} by associativity
− 2 + x = 4 {\displaystyle -2+x=4\,}
x = 4 + 2 {\displaystyle x=4+2\,}
x = 6 {\displaystyle x=6\,}
2 − x = 5 {\displaystyle 2-x=5\,}
− x = 3 {\displaystyle -x=3\,}
x = − 3 {\displaystyle x=-3\,} multiply both sides by -1.
Section 3
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30, 31, 32, 33
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Although these can be done by hand the exercises suggest that you derive a general formula for finding the final population figures. The following one will suffice for the stated problems, where p {\displaystyle p\,} x {\displaystyle x\,} y 1 {\displaystyle y_{1}\,} y 2 {\displaystyle y_{2}\,} n {\displaystyle n\,} x {\displaystyle x\,}
p ⋅ x ( y 2 − y 1 n ) {\displaystyle p\cdot x^{({\frac {y_{2}-y_{1}}{n}})}}
200000 ⋅ 3 ( 2215 − 1915 50 ) = 200000 ⋅ 3 6 = 145800000 {\displaystyle 200000\cdot 3^{({\frac {2215-1915}{50}})}=200000\cdot 3^{6}=145800000}
Section 4
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Write out the powers of 2 not exceeding the largest number in the problem set. Systematically check to see if each power divides the given number, starting with the largest.
Write out the powers of 3 not exceeding the largest number in the problem set. Systematically check to see if each power divides the given number, starting with the largest.
Examining the definitions first, if a ≡ b ( mod 5 ) {\displaystyle a\equiv b{\pmod {5}}} x ≡ y ( mod 5 ) {\displaystyle x\equiv y{\pmod {5}}} a − b = 5 n {\displaystyle a-b=5n\,} x − y = 5 m {\displaystyle x-y=5m\,}
a + x ≡ b + y ( mod 5 ) {\displaystyle a+x\equiv b+y{\pmod {5}}} ( a + x ) − ( b + y ) = 5 k {\displaystyle (a+x)-(b+y)=5k\,}
( a + x ) − ( b + y ) = ( a − b ) + ( x − y ) {\displaystyle (a+x)-(b+y)=(a-b)+(x-y)\,}
( a + x ) − ( b + y ) = 5 n + 5 m {\displaystyle (a+x)-(b+y)=5n+5m\,}
( a + x ) − ( b + y ) = 5 ( n + m ) {\displaystyle (a+x)-(b+y)=5(n+m)\,}
k = ( n + m ) {\displaystyle k=(n+m)\,}
( a + x ) − ( b + y ) = 5 k {\displaystyle (a+x)-(b+y)=5k\,}
Examine the definitions in part a) again. We have a − b = 5 n {\displaystyle a-b=5n\,} x − y = 5 m {\displaystyle x-y=5m\,}
a {\displaystyle a\,} x {\displaystyle x\,}
a = 5 n + b {\displaystyle a=5n+b\,} x = 5 m + y {\displaystyle x=5m+y\,}
a x = ( 5 n + b ) ( 5 m + y ) {\displaystyle ax=(5n+b)(5m+y)\,}
a x = 25 n m + 5 n y + 5 m b + b y {\displaystyle ax=25nm+5ny+5mb+by\,}
a x = 5 ( 5 n m + n y + m b ) + b y {\displaystyle ax=5(5nm+ny+mb)+by\,}
t = 5 n m + n y + m b {\displaystyle t=5nm+ny+mb\,}
a x = 5 t + b y {\displaystyle ax=5t+by\,}
a x − b y = 5 t + b y − b y {\displaystyle ax-by=5t+by-by\,}
a x − b y = 5 t {\displaystyle ax-by=5t\,}
Section 5
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120, 720, 5040 and 40320.
( m n ) = ( m m − n ) {\displaystyle {\binom {m}{n}}={\binom {m}{m-n}}}
m ! n ! ( m − n ) ! = m ! ( m − n ) ! ( m − ( m − n ) ) ! {\displaystyle {\frac {m!}{n!(m-n)!}}={\frac {m!}{(m-n)!(m-(m-n))!}}}
m ! n ! ( m − n ) ! = m ! ( m − n ) ! ( m − m + n ) ! {\displaystyle {\frac {m!}{n!(m-n)!}}={\frac {m!}{(m-n)!(m-m+n)!}}}
m ! n ! ( m − n ) ! = m ! ( m − n ) ! n ! {\displaystyle {\frac {m!}{n!(m-n)!}}={\frac {m!}{(m-n)!n!}}}
( m n ) + ( m n − 1 ) = ( m + 1 n ) {\displaystyle {\binom {m}{n}}+{\binom {m}{n-1}}={\binom {m+1}{n}}}
m ! n ! ( m − n ) ! + m ! ( n − 1 ) ! ( m − n + 1 ) ! = ( m + 1 ) ! n ! ( m − n + 1 ) ! {\displaystyle {\frac {m!}{n!(m-n)!}}+{\frac {m!}{(n-1)!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
n n {\displaystyle {\frac {n}{n}}} ( m − n + 1 ) ( m − n + 1 ) {\displaystyle {\frac {(m-n+1)}{(m-n+1)}}} n ! ( m − n + 1 ) ! {\displaystyle n!(m-n+1)!}
1 ⋅ ( m − n + 1 ) ( m − n + 1 ) ⋅ m ! n ! ( m − n ) ! + n n ⋅ 1 ⋅ m ! ( n − 1 ) ! ( m − n + 1 ) ! = 1 ⋅ 1 ⋅ ( m + 1 ) ! n ! ( m − n + 1 ) ! {\displaystyle 1\cdot {\frac {(m-n+1)}{(m-n+1)}}\cdot {\frac {m!}{n!(m-n)!}}+{\frac {n}{n}}\cdot 1\cdot {\frac {m!}{(n-1)!(m-n+1)!}}=1\cdot 1\cdot {\frac {(m+1)!}{n!(m-n+1)!}}}
a ! ( a + 1 ) = ( a + 1 ) ! {\displaystyle a!(a+1)=(a+1)!} a ( a − 1 ) ! = a ! {\displaystyle a(a-1)!=a!}
m ! ( m − n + 1 ) n ! ( m − n + 1 ) ! + m ! n n ! ( m − n + 1 ) ! = ( m + 1 ) ! n ! ( m − n + 1 ) ! {\displaystyle {\frac {m!(m-n+1)}{n!(m-n+1)!}}+{\frac {m!n}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
m ! ( m − n + 1 ) + m ! n n ! ( m − n + 1 ) ! = ( m + 1 ) ! n ! ( m − n + 1 ) ! {\displaystyle {\frac {m!(m-n+1)+m!n}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
m ! ( ( m − n + 1 ) + n ) n ! ( m − n + 1 ) ! = ( m + 1 ) ! n ! ( m − n + 1 ) ! {\displaystyle {\frac {m!((m-n+1)+n)}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
m ! ( m + 1 ) n ! ( m − n + 1 ) ! = ( m + 1 ) ! n ! ( m − n + 1 ) ! {\displaystyle {\frac {m!(m+1)}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
( m + 1 ) ! n ! ( m − n + 1 ) ! = ( m + 1 ) ! n ! ( m − n + 1 ) ! {\displaystyle {\frac {(m+1)!}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
( m n ) + ( m n − 1 ) = ( m + 1 n ) {\displaystyle {\binom {m}{n}}+{\binom {m}{n-1}}={\binom {m+1}{n}}}
Section 6
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2 x − 1 3 x + 2 = 7 {\displaystyle {\frac {2x-1}{3x+2}}=7\,}
2 x − 1 = 7 ( 3 x + 2 ) {\displaystyle 2x-1=7(3x+2)\,}
2 x − 1 = 21 x + 14 {\displaystyle 2x-1=21x+14\,}
− 19 x = 15 {\displaystyle -19x=15\,}
x = − 15 19 {\displaystyle x=-{\frac {15}{19}}\,}
1 x + y − 1 x − y = − 2 y x 2 − y 2 {\displaystyle {\frac {1}{x+y}}-{\frac {1}{x-y}}={\frac {-2y}{x^{2}-y^{2}}}}
( x − y ) − ( x + y ) ( x + y ) ( x − y ) = . . . {\displaystyle {\frac {(x-y)-(x+y)}{(x+y)(x-y)}}=...}
x + ( − x ) + ( − y ) + ( − y ) x 2 + y x − y x − y 2 = . . . {\displaystyle {\frac {x+(-x)+(-y)+(-y)}{x^{2}+yx-yx-y^{2}}}=...}
− 2 y x 2 − y 2 = − 2 y x 2 − y 2 {\displaystyle {\frac {-2y}{x^{2}-y^{2}}}={\frac {-2y}{x^{2}-y^{2}}}}
x 3 − 1 x − 1 = 1 + x + x 2 {\displaystyle {\frac {x^{3}-1}{x-1}}=1+x+x^{2}}
Recall from the chapter that if a b = c d {\displaystyle {\frac {a}{b}}={\frac {c}{d}}} a d = b c {\displaystyle ad=bc}
1 ⋅ ( x 3 − 1 ) = ( x − 1 ) ( 1 + x + x 2 ) {\displaystyle 1\cdot (x^{3}-1)=(x-1)(1+x+x^{2})}
x 3 − 1 = x + x 2 + x 3 − 1 − x − x 2 {\displaystyle x^{3}-1=x+x^{2}+x^{3}-1-x-x^{2}\,}
x 3 − 1 = x 3 − 1 {\displaystyle x^{3}-1=x^{3}-1\,}
