Download Sometimes / always / never

Sometimes / always / never?
Answers added in italics
For the following statements:
(a)
decide whether they are sometimes, always or never true;
(b)
if sometimes, when are they true?
(c)
work out how to persuade a friend you’re right;
(d)
work out how to persuade an enemy you’re right;
(e)
how would you describe the arguments you’ve used?
1.
Two consecutive integers add to make a multiple of
2.
Never true.
If the first integer is x, the next is x+1, so the total is 2x+1. 2x will
always be even, so 2x+1 will always be odd.
2.
Three consecutive integers add to make a multiple
of 3.
Always true.
If the first integer is x, the next is x+1, the next after that is x+2, the
total is 3x+3 = 3(x+1). x is an integer, so x+1 is an integer, so
3(x+1) will be a multiple of 3.
3.
n consecutive integers add to make a multiple of n.
Sometimes true: true when n is odd, false when n is even
[Need to know that 1+2+3+…..+n-1 = 1/2n(n-1)]
If the first integer is x, then the n consecutive integers starting at x
are: x, x+1,x+2, x+3, x+4……x+n-1.
The sum of these is
x + (x+1)+(x+2)+…..+(x+n-1)
= x+x+x+…+x + 1 + 2 + 3 + 4+…+ n-1
=nx + 1/2n(n-1).
nx will be a multiple of n, so need to know if 1/2n(n-1) is.
Case 1: n is odd.
Put 1/2n(n-1) = 1/2(n-1) x n
If n is odd, then n-1 is even, so 1/2(n-1) is an integer, so 1/2(n-1)n
is a multiple of n.
Case 2: n is even
If n is even, then n-1 is odd, so 1/2(n-1) is not an integer, so
1/2(n-1)n is not a multiple of n.
4.
Two consecutive integers multiply to make a
multiple of 2.
Always true.
For any two consecutive integers, one will be odd, one will be
even. When an odd is multiplied to an even, you get an even
number.
5.
Three consecutive integers multiply to make a
multiple of 3.
Always true.
Any three consecutive integers will contain a multiple of 3. So the
three numbers multiplied together will have a multiple of 3.
6.
n consecutive integers multiply to make a multiple of
n.
Always true.
Any n consecutive integers will contain a multiple of n. So the
numbers multiplied together will have a multiple of n.
7.
Two consecutive odd integers add to make a
multiple of 4.
Always true.
If x is an integer, then 2x+1 is odd, 2x+3 is the next odd number,
add them together, get 4x+4 = 4(x+1) which is a multiple of 4.
8.
All prime numbers are odd.
Not true: 2 is the only even prime
A prime is a number with only itself and 1 as factors. True of 2.
Any other even number will have 2 as a factor, so not a prime.
9.
It is possible to find the biggest prime number.
Not true: there are infinitely many primes.
Suppose that we have the complete list of primes, p1, p2, p3…pn.
Suppose that we multiply all these primes together and call the
answer M. So M has a factor of p1, p2, p3….pn. But if p1 is a factor
of M, it cannot be a factor of M+1. Similarly M+1 cannot have a
factor of any of p1, p2, p3…pn. So either M+1 is prime, or it has a
completely different set of prime factors to M. Either way we have
found further primes, disproving the original assertion of a
complete list of primes.
10.
x2 > x.
Sometimes true: not true for 0 ≤ x ≤ 1.
Approach 1: draw a graph
5
4
3
2
1
0
-1
-1
-0.5
0
0.5
1
1.5
2
-2
Approach 2: using inequalities
If
x2 > x
then
x2 – x > 0
so
x(x-1)> 0
So either x and x-1 both >0, ie. x>1
Or
x and x-1 both <0, ie. x<0.
11.
m ÷ n ≤ m if m and n are positive integers.
Always true.
If n = 1, then m ÷ n = m. If n > 1, then m ÷ n < m.
12.
m ÷ n ≤ m if m and n can be any numbers.
Sometimes true.
Case 1: m and n both positive: false for n<1, otherwise true.
Case 2: m negative, n positive: true for n<1, otherwise false.
Case 3: m positive, n negative: necessarily true.
Case 4: m and n both negative: necessarily false.
13.
x2 – 1 = (x+1)(x-1).
Always true. See appendix 1.
14.
starting at 1, if you add together n odd numbers, you
will get a square number, where n is any positive
integer.
Always true.
Equivalent to saying that x2 = 1 + 3 + 5 + 7 +… + 2x-1
Approach 1
So need x2 = (x-1)2 + (2x-1)
= x2 –2x + 1 +2x –1
= x2 as required
Approach 2
1 + 3 + 5 + 7 + 9 + … + 2x-1
=
(0+1) + (2+1) + (4+1) +… + (2x-2+1)
=
(0 + 2 +_4 + 6 + …. + 2x-2) + (1+1+1+1+1+...+1)
=
2(1+2+3+4+…+x-1) +x
=
2 . ½ . x.(x-1) + x
=
x2 – x + x
=
x2 as required.
Approach 3
By diagram:
x-1
x
x
x-1
x
x
Square in grey added twice, so adding 2x-1 to get next square.
15.
If you add 1, ½, ¼, 1/8, 1/16 and so on, you will
never actually get to 2.
Always true.
Approach 1: by diagram
Approach 2: by using the formula for a geometric progression:
1+r+r2+r3+…+rn-1 = (1-rn)/(1-r) for r<1
1+1/2+1/4+1/16+….+1/2n-1
= (1-(1/2)n)/(1-1/2)
= 2(1-(1/2)n)
< 2 for any value of n.
16.
n2 + n + 41 will give you a prime number if n is a
positive integer or 0.
Sometimes true.
Appears initially to be true, but clearly not when n = 41.
17.
 b  (b 2  4ac)
If ax + bx + c = 0, then x =
2a
2
Always true
ax2 + bx + c
=0
x2 + bx/a
= -c/a
(x+ b/2a)2
= b2/4a2 –c/a
[Completing the square; we’ll
= (b2-4ac)/4a2
come back to this next time]
x+b/2a
= ± √(b2-4ac)/2a
x
=
 b  (b 2  4ac)
2a
This solution is always true so long as we allow complex solutions.
Otherwise it’s sometimes true for b2 – 4ac ≥ 0.
18.
the square root of a whole number is irrational.
Sometimes true: false when the whole number is a perfect square,
otherwise true.
Take 2 as an example, and assume this is false. So 2 is rational,
ie. can be written as m/n where m and n are coprime, ie. have no
common factors.
So m/n = 2
m2/n2 = 2
m2 = 2n2
Therefore m2 is even, which means that m must be even. Write m
= 2p
2p/n = 2
p/n = 2 / 2
p/n = 1 / 2
n/p = 2
n2/p2 = 2
n2 = 2p2
Therefore n2 is even, which means that n must be even.
But if both n and m are even, this contradicts the assertion that n
and m are coprime.
So original proposition must be correct.
[You may like to check that this argument works for 3. Why does it
fail for 4?]
19.
(a+b)2 = a2 + b2
Sometimes true, ie. when a or b = 0, otherwise false.
In general (a+b)2 = a2 +2ab + b2
a
Approach 1: by diagram:
b
a
a
b
b
a
Approach 2: by algebraic manipulation
(a+b)2
= a2 + ab + ab + b2
= a2 +2ab + b2
b
Appendix 1:
x2 – 1 = (x+1)(x-1)
Approach 1: visual
Introductory Example
x
x
x -1
x
x–1
x
x -1
x–1
x
x–1
x
x–1
x +1
Approach 2: by example
x
1
2
3
4
5
6
7
8
9
10
100
1000
0
1.5
7.7948
-6.4
x2
x2 -1
1
0
4
3
9
8
16
15
25
24
36
35
49
48
64
63
81
80
100
99
10000
9999
1000000 999999
0
-1
2.25
1.25
60.75891 59.75891
40.96
39.96
x-1
0
1
2
3
4
5
6
7
8
9
99
999
-1
0.5
6.7948
-7.4
x+1
2
3
4
5
6
7
8
9
10
11
101
1001
1
2.5
8.7948
-5.4
(x-1)(x+1)
0
3
8
15
24
35
48
63
80
99
9999
999999
-1
1.25
59.75891
39.96
Ie. not a proof, but pretty convincing evidence!
Approach 3: by diagram showing y = x2 against y = (x+1)(x-1)
10
8
6
4
2
-2
3
-3
-2
.4
-1
.8
-1
.2
1. 52 0.
7E 6
-1
5
0.
6
1.
2
1.
8
2.
4
0
BUT: data comes from table above, so no more of a proof than a
table is. Also, does this diagram convince you that the graphs are
always 1 apart? If not, why not?
Approach 4: formal algebraic manipulation
(x+1)(x-1) =
x2 – x + x –1
=
x2 – 1
BUT whilst this covers every case, how accessible is it to children?