The Student Room Group

The Proof is Trivial!


I noticed that the mathematics forum is currently lacking a thread where people can post interesting and challenging problems for fellow mathematicians to ponder, so here it is.

Please don't post here for queries on A level or STEP questions, as they each have their respective threads. The level of the questions can of course vary, but they should be fun, interesting, and most importantly solvable! Please don't post questions that you're not able or willing to provide solutions to. Questions that require knowledge of graduate level theorems won't be of interest to most of us, so keep the level reasonable.



Use the following format for problems and solutions:

(example)



Problem 0*

Show that for any nn there exist nn consecutive integers such that none of them are prime.



Solution 0

(n+1)!+2,  (n+1)!+3,    (n+1)!+(n+1)(n+1)!+2,\;(n+1)!+3,\;\cdots\; (n+1)!+(n+1)

To avoid posts such as "what do I need to know to solve this?", use one of the following tags with your question (see above):

* = requires only A-level knowledge
** = may require a little extra (induction, L'hôpital's etc.)
*** = requires undergraduate knowledge

(the stars do not reflect the difficulty of the question)


Please LaTex all problems and solutions, and remember to spoiler any hints.



Below is a set of links to the problems/solutions posted in the thread. Please remember to number your problem accordingly.

Problem 1 · Solution (electriic_ink)
Problem 2 · Solution (Lord of the Flies)
Problem 3 · Solution (Mladenov)
Problem 4 · Solution (Lord of the Flies)
Problem 5 · Solution (Felix Felicis)
Problem 6 · Solution (Mladenov)
Problem 7 · Solution (j.alexanderh)
Problem 8 · Solution (Lord of the Flies)
Problem 9 · Solution (Noble.)
Problem 10 · Solution (Lord of the Flies)
Problem 11 · Solution (Smaug123)
Problem 12 · Solution (und) · Solution 2 (LLewellyn)
Problem 13 · Solution (Lord of the Flies)
Problem 14 · Solution (Noble.) · Solution 2 (metaltron) · Solution 3 (Hasufel)
Problem 15 · Solution (_Izzy)
Problem 16 · Solution (Farhan.Hanif93)
Problem 17 · Solution (und)
Problem 18 · Solution (Smaug123)
Problem 19 · Solution (und)
Problem 20 · Solution (Felix Felicis)
Problem 21 · Solution (und)
Problem 22 · Solution (und)
Problem 23 · Solution (Mladenov)
Problem 24 · Solution (Mladenov)
Problem 25 · Solution (Mladenov) Comments
Problem 26 · Solution (Mladenov)
Problem 27 · Solution to first part (Star-girl)
Problem 28 · Solution (Felix Felicis)
Problem 29 · Solution (Lord of the Flies)
Problem 30 · Solution (und)
Problem 31 · Solution (Upper Echelons)
Problem 32 · Solution (Felix Felicis)
Problem 33 · Solutions 1 & 2 (Lord of the Flies) · Solution 3 (Star-girl) Solution 4 (Felix Felicis)
Problem 34 · Solution (DJMayes) · Solution 2 (metaltron)
Problem 35 · Solution (Felix Felicis)
Problem 36 · Solution (Felix Felicis)
Problem 37 · Solution (Lord of the Flies)
Problem 38 · Solution (Mladenov)
Problem 39
Problem 40 · Solution (Lord of the Flies)
Problem 41 · Solution (Mladenov)
Problem 42 · Solution (Noble.)
Problem 43 · Solution (und)
Problem 44 · Solution (jack.hadamard) · Comments
Problem 45 · Solution (und)
Problem 46 · Solution (DJMayes)
Problem 47 · Solution (Lord of the Flies)
Problem 48 · Solution (Mladenov)
Problem 49 · Solution (Mladenov)
Problem 50 · Solution (Star-girl) · Solution 2 (DJMayes) · Solution 3 (und)
Problem 51
Problem 52 · Solution (Lord of the Flies)
Problem 53 · Solution (Lord of the Flies)
Problem 54 · Solution (metaltron) · Comments · Comments
Problem 55 · Solution (Llewellyn)
Problem 56 · Solution (Felix Felicis)
Problem 57 · Solution (Farhan.Hanif93)
Problem 58 · Solution (bananarama2)
Problem 59 · Solution (Mladenov)
Problem 60 · Solution (Lord of the Flies)
Problem 61 · Solution (Lord of the Flies)
Problem 62 · Solution (Mladenov)
Problem 63 · Solution (bananarama2)
Problem 64 · Solution (Mladenov) · Comments
Problem 65
Problem 66 · Solution (Mladenov)
Problem 67 · Solution (Mladenov)
Problem 68 · Solution (Lord of the Flies)
Problem 69 · Solution (Mladenov) · Solution 2 (ben-smith)
Problem 70 · Solution (Mladenov)
Problem 71 · Solution (james22)
Problem 72 · Solution (Indeterminate)
Problem 73 · Solution (Mladenov)
Problem 74
Problem 75 · Solution (Mladenov)
Problem 76 · Solution (Mladenov) · Solution 2 (Jkn) · Solution 3 (Jkn) · Solution 4 (Jkn)
Problem 77 · Solution (Mladenov)
Problem 78 · Solution (Mladenov)
Problem 79 · Solution (Mladenov)
Problem 80 · Solution (Lord of the Flies) · Solution 2 (Jkn) · Solution 3 (Zephyr1011)
Problem 81
Problem 82
Problem 83
Problem 84
Problem 85 · Solution (Jkn)
Problem 86 · Solution (Jkn) · Solution 2 (james22)
Problem 87 · Solution (MW24595)
Problem 88 · Solution (Mladenov) · Solution 2 (Zephyr1011) · Solution 3 (Jkn)
Problem 89 · Solution (Mladenov)
Problem 90 · Solution (metaltron) · Solution to second part (Jkn)
Problem 91 · Solution (Jkn)
Problem 92 · Solution (Lord of the Flies) · Solution 2 (Jkn)
Problem 93 · Solution (Mladenov) · Comments
Problem 94 · Solution (Zephyr1011) · Solution 2 (Lord of the Flies)
Problem 95 · Solution (Jkn)
Problem 96
Problem 97 · Solution (Lord of the Flies)
Problem 98 · Solution (Lord of the Flies)
Problem 99 · Solution (Lord of the Flies)
Problem 100 · Solution (Mladenov)
Problem 101 · Solution (Mladenov)
Problem 102 · Solution (Mladenov)
Problem 103 · Solution (ben-smith) · Solution 2 (Lord of the Flies)
Problem 104 · Solution (Mladenov)
Problem 105 · Solution (Mladenov)
Problem 106 · Solution (Lord of the Flies) · Solution 2 (Jkn)
Problem 107 · Solution (Lord of the Flies)
Problem 108 · Solution (und) · Solution 2 (Lord of the Flies)
Problem 109 · Solution (Lord of the Flies)
Problem 110 · Solution (FireGarden)
Problem 111 · Solution (Mladenov)
Problem 112 · Solution (Lord of the Flies)
Problem 113
Problem 114 · Solution (Lord of the Flies)
Problem 115 · Solution (Lord of the Flies)
Problem 116
Problem 117 · Solution (Lord of the Flies)
Problem 118
Problem 119 · Solution (Mladenov) · Solution 2 (Lord of the Flies)
Problem 120 · Solution (Lord of the Flies)
Problem 121 · Solution (Lord of the Flies)
Problem 122
Problem 123 · Solution (Mladenov)
Problem 124 · Solution (Mladenov)
Problem 125
Problem 126 · Solution (ben-smith)
Problem 127 · Solution (Mladenov)
Problem 128 · Solution (bananarama2)
Problem 129 · Solution (bananarama2)
Problem 130 · Solution (Lord of the Flies)
Problem 131 · Solution (ben-smith)
Problem 132 · Solution (Felix Felicis) · Solution 2 (ben-smith)
Problem 133 · Solution (Felix Felicis) · Solution 2 (Mladenov)
Problem 134 · Solution (ben-smith)
Problem 135 · Solution (Mladenov)
Problem 136
Problem 137 · Solution (Lord of the Flies)
Problem 138
Problem 139 · Solution (Mladenov) · Extension · Solution to extension (Mladenov)
Problem 140 · Solution (Felix Felicis) · Solution 2 (Lord of the Flies) · Solution 3 (Mladenov)
Problem 141
Problem 142 · Solution (Lord of the Flies)
Problem 143
Problem 144
Problem 145 · Solution (jack.hadamard)
Problem 146 · Solution (Mladenov)
Problem 147 · Solution (Mladenov)
Problem 148 · Solution (Lord of the Flies)
Problem 149
Problem 150 · Solution (bananarama2)
Problem 151
Problem 152
Problem 153
Problem 154 · Solution (Jkn)
Problem 155
(cont. in post 2)

(N.B. If a solution has been posted but you wish to add your own contribution, please feel free; both will be linked to in this post.)
(edited 10 years ago)

Scroll to see replies

Reply 1
Problem 156 · Solution (Mladenov) · Solution 2 (ben-smith) · Solution 3 (Jkn)
Problem 157 · Solution (Mladenov) · Solution 2 (Lord of the Flies)
Problem 158 · Solution (Mladenov) · Solution 2 (Lord of the Flies)
Problem 159 · Solution (Mladenov)
Problem 160 · Solution (Lord of the Flies)
Problem 161
Problem 162 · Hint
Problem 163 · Solution (Lord of the Flies)
Problem 164 · Solution (Lord of the Flies)
Problem 165
Problem 166 · Solution (Lord of the Flies)
Problem 167
Problem 168 · Solution (Mladenov) · Solution 2 (Felix Felicis)
Problem 169 · Solution (Lord of the Flies) · Solution 2 (Jkn)
Problem 170 · Solution (Lord of the Flies)
Problem 171 · Solution 1 (DJMayes) · Solution 2 (Zakee)
Problem 172 · Solution to first part (DJMayes) · Solution to second part (Zakee)
Problem 173 · Solution (Lord of the Flies)
Problem 174 · Solution (Felix Felicis) · Solution 2 (Jkn)
Problem 175 · Solution (Felix Felicis) Solution 2 (DJMayes)
Problem 176 · Solution (joostan)
Problem 177 · Solution (Felix Felicis)
Problem 178 · Solution (DJMayes) · Solution 2 (Jkn)
Problem 179 · Solution (joostan)
Problem 180 · Solution (Felix Felicis)
Problem 181
Problem 182
Problem 183 · Solution (joostan) · Solution 2 (Jkn)
Problem 184 · Solution (DJMayes) · Solution 2 (LewisMead)
Problem 185 · Solution (Mladenov)
Problem 186 · Solution (Mladenov)
Problem 187 · Solution (Lord of the Flies)
Problem 188 · Solution (Jkn)
Problem 189 · Solution (Benjy100)
Problem 190 · Solution (Felix Felicis) · Solution 2 (such) · Solution 3 (LewisMead)
Problem 191 · Solution (Felix Felicis)
Problem 192 · Solution (Mladenov)
Problem 193 · Solution (Mladenov)
Problem 194
Problem 195 · Solution (DJMayes) · Solution 2 (Mladenov)
Problem 196 · Solution (Mladenov)
Problem 197 · Solution (metaltron)
Problem 198
Problem 199
Problem 200 · Solution (Felix Felicis) · Problem 200 · Solution (jack.hadamard)
Problem 201 · Solution (bananarama2)
Problem 202 · Solution (DJMayes) · Problem 202
Problem 203
Problem 204
Problem 205
Problem 206 · Solution (bananarama2)
Problem 207
Problem 208
Problem 209 · Solution (MW24595)
Problem 210 · Solution (MW24595) Solution 2 (Jkn)
Problem 211 · Solution (Lord of the Flies)
Problem 212 · Solution (Jkn) · Solution 2 (Lord of the Flies)
Problem 213 · Solution (Mladenov) · Solution 2 (Lord of the Flies)
Problem 214 · Solution (Lord of the Flies)
Problem 215 · Solution (Lord of the Flies)
Problem 216 · Solution (Lord of the Flies)
Problem 217 · Solution (Lord of the Flies)
Problem 218 · Solution (joostan)
Problem 219 · Solution (ukdragon37)
Problem 220
Problem 221 · Solution (Lord of the Flies)
Problem 222 · Solution (Jkn)
Problem 223 · Solution (Mladenov)
Problem 224 · Solution (Lord of the Flies)
Problem 225
Problem 227
Problem 228
Problem 229 · Solution (Mladenov)
Problem 230 · Solution (Lord of the Flies) · Solution 2 (ukdragon37)
Problem 231 · Solution (Mladenov) · Solution 2 (Jkn)
Problem 232 · Solution (Mladenov)
Problem 233 · Solution (Mladenov) · Solution 2 (Jkn) · Solution 3 (Jkn)
Problem 234
Problem 235 · Solution (Mladenov)
Problem 236 · Solution (The Polymath) · Solution 2 (henpen)
Problem 237 · Solution (Felix Felicis)
Problem 238 · Solution (The Polymath)
Problem 239 · Solution (james22)
Problem 240 · Solution (Lord of the Flies)
Problem 241 · Solution (TheMagicMan) · Solution 2 (Jkn)
Problem 242
Problem 243 · Solution (Ateo)
Problem 244
Problem 245 · Solution (Mladenov)
Problem 246 · Solution (Mladenov) · Solution 2 (Lord of the Flies)
Problem 247 · Solution (FireGarden)
Problem 248
Problem 249
Problem 250 · Solution (Mladenov) · Solution 2 (Blutooth)
Problem 251 · Solution (jack.hadamard)
Problem 252
Problem 253
Problem 254
Problem 255
Problem 256

(N.B. If a solution has been posted but you wish to add your own contribution, please feel free; both will be linked to in this post.)




Problem 1*

Prove that if the function f\displaystyle f is continuous on [a,b]\displaystyle [a,b] and abf(x)g(x)dx=0\displaystyle \int^b_af(x)g(x) dx = 0 for any continuous function g\displaystyle g, then f=0\displaystyle f=0 on [a,b]\displaystyle [a,b].
(edited 10 years ago)
Original post by und
Problem 1*

Prove that if the function f\displaystyle f is continuous on [a,b]\displaystyle [a,b] and abf(x)g(x)dx=0\displaystyle \int^b_af(x)g(x) dx = 0 for any continuous function g\displaystyle g, then f=0\displaystyle f=0 on [a,b]\displaystyle [a,b].


Solution 1 (attempt)

Spoiler

(edited 10 years ago)
I'll the post the next problem. Undergraduates will probably find it pretty easy and it can be done with GCSE knowledge.

Problem 2*

Define a function f:R R using the following rule:

"For any x,y∈R, f(x+y)=f(x)f(y)"

Show that, if f is non-zero:

(a) f(0)=1
(b) f(x)=0 has no solutions
(c) f(-x)=1/f(x)
(edited 10 years ago)
Solution 2

f(x)=f(0)f(x)f(0)=1f(x)=f(0)f(x)\Rightarrow f(0)=1. Suppose ff has a solution cc then f(x+c)=f(c)f(x)=0f=0f(x+c)=f(c)f(x)=0\Rightarrow f=0, contradiction. Set y=x:  f(0)=f(x)f(x)f(x)=f(x)1y = -x:\; f(0)=f(x)f(-x)\Rightarrow f(-x)=f(x)^{-1}
(edited 10 years ago)
Problem 3*

Show that for positive integers m,n:m,n:

011xnm1xmndx=0\displaystyle\int_0^1 \sqrt[m]{1-x^n}-\sqrt[n]{1-x^m}\,dx=0
(edited 10 years ago)
I'm getting too old for this :coffee:

Spoiler

(edited 10 years ago)
Reply 7
solution 2

f(x + 0) = f(x)f(0)

f(x) = f(x)f(0)

=> f(0) = 1
Reply 8
Problem 4*

If N is any perfect power of 2 (i.e. N=2^n for some integer n) then there is no number, M, formed by permutating the digits of N, which is also a perfect power of 2.
Original post by und
Problem 1*

Prove that if the function f\displaystyle f is continuous on [a,b]\displaystyle [a,b] and abf(x)g(x)dx=0\displaystyle \int^b_af(x)g(x) dx = 0 for any continuous function g\displaystyle g, then f=0\displaystyle f=0 on [a,b]\displaystyle [a,b].


That's not true, otherwise there would not exist non-trivial orthogonal functions. For a concrete example;

ππsin(x)sin(2x)dx=0 \displaystyle\int_{-\pi}^{\pi} sin(x)sin(2x) dx = 0

even though all the criteria specified hold here.


EDIT: In fact, we don't even need to look to orthogonal functions. Take ff to be either odd or even, then choose gg to be of opposite parity. Then the product fgfg is odd. Take the interval [a,a] [-a,a] and the integral of fgfg shall be zero as it is odd, and integrated over symmetric limits.
(edited 10 years ago)
Reply 10
Original post by FireGarden
That's not true, otherwise there would not exist non-trivial orthogonal functions. For a concrete example;

ππsin(x)sin(2x)dx=0 \displaystyle\int_{-\pi}^{\pi} sin(x)sin(2x) dx = 0

even though all the criteria specified hold here.

For any continuous function g, not just one example of your choice.
Reply 11
Problem 5** De Moivre's Theorem

By considering the real and imaginary parts of:
 eiθeiϕ\ e^{i \theta}\cdot e^{i \phi}
Prove that sin(θ+ϕ)=sin(θ)cos(ϕ)+sin(ϕ)cos(θ)\sin(\theta + \phi) = sin(\theta) cos(\phi) + sin(\phi) cos(\theta)
and that cos(θ+ϕ)=cos(θ)cos(ϕ)sin(θ)sin(ϕ)\cos(\theta + \phi) = cos(\theta) cos(\phi) - sin(\theta) sin(\phi)


Show also that:
 S1(n)=nC0nC2+nC4...+(1)m×nC2m has the value 2n2cos(nπ4)\ S_1 (n) = ^n\mathrm{C}_0 - ^n\mathrm{C}_2 + ^n\mathrm{C}_4 - . . . +(-1)^m \times ^n\mathrm{C}_{2m} \ \mathrm{has\ the\ value}\ 2^{\frac{n}{2}} cos(\frac{n\pi}{4})
Where  n12mn\ {n - 1} \leq{2m} \leq{n}

Derive another expression for:
 S2(n)=nC1nC3+nC5...+(1)m×nC2m+1\ S_2 (n) = ^n\mathrm{C}_1 - ^n\mathrm{C}_3 + ^n\mathrm{C}_5 - . . . +(-1)^m \times ^n\mathrm{C}_{2m+1} Where  n12m+1n\ n - 1 \leq 2m+1 \leq n
(edited 10 years ago)
Original post by ghostwalker
It's not true, in general.


Are you sure?
Reply 13
Original post by joostan
5**

By considering the real and imaginary parts of:
 eiθeiϕ\ e^{i \theta}\cdot e^{i \phi}
Prove that sin(θ+ϕ)=sin(θ)cos(ϕ)+sin(ϕ)cos(θ)\sin(\theta + \phi) = sin(\theta) cos(\phi) + sin(\phi) cos(\theta)
and that cos(θ+ϕ)=cos(θ)cos(ϕ)sin(θ)sin(ϕ)\cos(\theta + \phi) = cos(\theta) cos(\phi) - sin(\theta) sin(\phi)

This doesn't seem particularly fun, interesting or challenging. It looks like it's straight out of a standard A level textbook exercise!
Reply 14
Original post by und
This doesn't seem particularly fun, interesting or challenging. It looks like it's straight out of a standard A level textbook exercise!


I'm typing up the second part of the problem. I accidentally submitted before I finished.
Original post by Mr M
Are you sure?


Oops! Had +'s under the root signs, rather than -'s.

Thanks for spotting my error.
Reply 16
Original post by joostan
I'm typing up the second part of the problem. I accidentally submitted before I finished.

Is that a STEP question?
Reply 17
Original post by und
Is that a STEP question?


Not as far as I'm aware . . . It's derived from a textbook for Physics and Engineering, but is manageable with a hint for STEP candidates.
(edited 10 years ago)
Solution 4

Suppose their exist distinct integers n, m such that 2^n and 2^m satisfy the desired property. Then:

2n2m(mod9)2^n \equiv 2^m\pmod{9} (a number is always congruent to its digit sum mod 9)

However, this implies n = 6k+m which in turn implies that 2^n and 2^m cannot have the same number of digits. Hence no such n, m exist.
As expected, the problems I'm interested in are more quaint than the usual :tongue: Here's one that's strange but rather easy once you get it.

Problem 6**/***

Let Nω=N{ω}\mathbb{N}_\omega = \mathbb{N} \cup \{\omega\} where we define for all nNωn \in \mathbb{N}_\omega, ωn\omega \geq n. For any set SNωS \subseteq \mathbb{N}_\omega we write S\sqcup S to denote the smallest member of Nω\mathbb{N}_\omega such that Sn\sqcup S \geq n for all nSn \in S. (Hence Nω=ω\sqcup\mathbb{N}_\omega = \omega.)

Let f:NωNωf:\mathbb{N}_\omega \to \mathbb{N}_\omega be a function such that:

For all m, n, mn    f(m)f(n)m \leq n \implies f(m) \leq f (n).

For all S={a1,a2,}NωS = \{a_1, a_2, \dots\} \subseteq \mathbb{N}_\omega, we have f(S)={f(a1),f(a2),}f(\sqcup S) = \sqcup\{f(a_1), f(a_2), \dots\}.



Show that the value F given by F={fn(1)nN{0}}F = \sqcup\left\{f^n(1)|n \in \mathbb{N}\cup\{0\}\right\} satisfies f(F)=Ff(F) = F and is the smallest member of Nω\mathbb{N}_\omega to do so.

EDIT: Right I see where people are getting confused. ω\omega is not a variable - it is actually an element in addition to the naturals that is defined to be greater than or equal to all of them. See it as an element representing "infinity".
(edited 10 years ago)

Quick Reply

Latest