Bogus Proof Marathon

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#1 h Mar 12, 2018, 7:33 PM • 240 Y

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Hi!
I'd like to introduce the Bogus Proof Marathon.

In this marathon, simply post a bogus proof that is middle-school level and the next person will find the error. You don't have to post the real solution

Use classic Marathon format:

[hide=P#]a1b2c3[/hide]
[hide=S#]a1b2c3[/hide]

Example posts:

P(x)

S(x)
P(x+1)

Let's go!! Just don't make it too hard!

This post has been edited 3 times. Last edited by pifinity, Mar 12, 2018, 7:40 PM

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j3370

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j3370

#2 h Mar 12, 2018, 7:36 PM • 72 Y

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P1

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pifinity

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pifinity

#5 h Mar 12, 2018, 7:40 PM • 46 Y

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S1

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Math-Ninja

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Math-Ninja

#6 h Mar 12, 2018, 7:41 PM • 59 Y

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P3

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pifinity

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pifinity

#7 h Mar 12, 2018, 7:43 PM • 33 Y

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S2
P3

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Problemsolving17

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Problemsolving17

#8 h Mar 12, 2018, 7:44 PM • 33 Y

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Math-Ninja wrote:

P3

Isn't this supposed to be P2 not P3

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pifinity

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pifinity

#9 h Mar 12, 2018, 7:45 PM • 25 Y

Y by math_swim, Jc426, snow52, RedFireTruck, ImSh95, samrocksnature, Lilathebee, Mathdreams, RedFlame2112, son7, HWenslawski, ryusei, RhinocerosHornbill, 953324, ryanc226, jmiao, ZtotheV, ethancui0529, ShrewdBunny, lpieleanu, Adventure10, Spiritpalm, tacocatz, MathNinja9, Exponent11

Problemsolving17 wrote:

Math-Ninja wrote:

P3

Isn't this supposed to be P2 not P3

yeah

pifinity wrote:

S2
P3

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mathmonster21

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mathmonster21

#10 h Mar 12, 2018, 7:46 PM • 44 Y

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How is $2+0=1$ ?

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pifinity

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pifinity

#11 h Mar 12, 2018, 7:47 PM • 24 Y

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mathmonster21 wrote:

S3

FTFY

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Problemsolving17

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Problemsolving17

#12 h Mar 12, 2018, 7:47 PM • 23 Y

Y by math_swim, Jc426, fluffyclouds, snow52, RedFireTruck, samrocksnature, Lilathebee, Mathdreams, RedFlame2112, son7, innout, HWenslawski, ImSh95, ryusei, RhinocerosHornbill, jmiao, ethancui0529, lpieleanu, Adventure10, Mango247, Spiritpalm, tacocatz, MathNinja9

mathmonster21 wrote:

How is $2+0=1$ ?

pifinity wrote:

Hi!
I'd like to introduce the Bogus Proof Marathon.

In this marathon, simply post a bogus proof that is middle-school level and the next person will find the error. You don't have to post the real solution

Use classic Marathon format:

[hide=P#]a1b2c3[/hide]
[hide=S#]a1b2c3[/hide]

Example posts:

P(x)

S(x)
P(x+1)

Let's go!! Just don't make it too hard!

Z K Y

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Problemsolving17

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Problemsolving17

#13 h Mar 12, 2018, 7:49 PM • 25 Y

Y by math_swim, darkwater, Jc426, snow52, Pi-rate, RedFireTruck, samrocksnature, RedFlame2112, Lilathebee, Mathdreams, hdrcure, son7, innout, HWenslawski, ImSh95, ryusei, RhinocerosHornbill, ryanc226, jmiao, ethancui0529, lpieleanu, Adventure10, Mango247, michael5210, MathNinja9

S3

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pifinity

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pifinity

#14 h Mar 12, 2018, 7:49 PM • 20 Y

Y by math_swim, snow52, RedFireTruck, samrocksnature, RedFlame2112, Lilathebee, Mathdreams, son7, innout, HWenslawski, ImSh95, ryusei, RhinocerosHornbill, jmiao, ethancui0529, lpieleanu, Adventure10, Mango247, Spiritpalm, MathNinja9

pifinity wrote:

mathmonster21 wrote:

S3

FTFY

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pifinity

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pifinity

#15 h Mar 12, 2018, 7:52 PM • 26 Y

Y by math_swim, cw357, BananaeGod, snow52, Pi-rate, RedFireTruck, samrocksnature, Lilathebee, Mathdreams, Rab7532, RedFlame2112, son7, innout, megarnie, HWenslawski, ImSh95, ryusei, RhinocerosHornbill, jmiao, ethancui0529, lpieleanu, Adventure10, Mango247, Spiritpalm, Easondragon, MathNinja9

P4

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Math-Ninja

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Math-Ninja

#16 h Mar 12, 2018, 7:55 PM • 23 Y

Y by math_swim, DRAGIC, snow52, RedFireTruck, samrocksnature, Mathdreams, RedFlame2112, son7, innout, megarnie, HWenslawski, ImSh95, ryusei, RhinocerosHornbill, jmiao, ethancui0529, lpieleanu, yompapike, Adventure10, Mango247, Spiritpalm, Yrock, MathNinja9

P5

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Problemsolving17

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Problemsolving17

#17 h Mar 12, 2018, 7:57 PM • 19 Y

Y by math_swim, snow52, ImSh95, RedFireTruck, samrocksnature, Mathdreams, RedFlame2112, son7, innout, HWenslawski, ryusei, RhinocerosHornbill, jmiao, ethancui0529, lpieleanu, Adventure10, Mango247, Spiritpalm, MathNinja9

S4

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e_is_2.71828

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e_is_2.71828

#8173 h Apr 21, 2025, 1:54 AM

Y by

iwastedmyusername wrote:

e_is_2.71828 wrote:

Pengu14 wrote:

Both solutions are wrong.

Oshawoot wrote:

S1973

Actually, your claim that the solutions are wrong is wrong. The issue is that in claiming $n$ to be the greatest positive integer and so (technically just greater than 1 lol), so $\mathbb{N}$ is thus unbounded. So both proofs are actually correct, and Oshawoot's solution is the incorrect one, as it fails to prove why the set of natural numbers is not bounded.

Actually, your claim that the solutions are wrong is wrong is wrong. You never disproved the claim $n^2 \le n$ , you simply stated that $n^2 \ge n$ . It's like using this logic to disprove P1974: "Well, no, pi can't be equal to 4, because pi is around 3.14."

What??? That is absurd to me that you actually even posted that. Equality is the only possible way both statements can be true, but besides that you just said "Oh, you just proved ~ $p$ is true not $p$ is false." Like what???

This post has been edited 1 time. Last edited by e_is_2.71828, Apr 21, 2025, 1:54 AM
Reason: LaTeX

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iwastedmyusername

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iwastedmyusername

#8174 h Apr 21, 2025, 2:18 AM • 1 Y

Y by e_is_2.71828

ok admittedly i just wanted to use the "your claim that the solutions are wrong is wrong is wrong."

I think the real issue with your solution is that you fail to address the root of the bogus proof, that there is no largest real number (which can easily be proven by n+1>n for a supposed largest real number n). If I said "This proof is bogus because 2, a real number, is greater than 1", would that be a valid solution?

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Pengu14

577 posts

Pengu14

#8175 h Apr 21, 2025, 9:35 AM

Y by

e_is_2.71828 wrote:

Pengu14 wrote:

Both solutions are wrong.

Oshawoot wrote:

S1973

Actually, your claim that the solutions are wrong is wrong. The issue is that in claiming $n$ to be the greatest positive integer, and so $n^2 \leq n$ , you are contradicting the proposition I proved, which is why your solution is wrong. In fact, an easy proof that $\mathbb{N}$ is unbounded is by letting $f(n)=n^2$ , and $f(n_i) \in \mathbb{N}$ for all $n_i \in \mathbb{N}$ , and $n_i < f(n_i)$ eventually for large enough $n_i$ (technically just greater than 1 lol), so $\mathbb{N}$ is thus unbounded. So both proofs are actually correct, and Oshawoot's solution is the incorrect one, as it fails to prove why the set of natural numbers is not bounded.

Did bro even read my “proof” :rotfl:

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e_is_2.71828

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e_is_2.71828

#8176 h Apr 21, 2025, 12:38 PM

Y by

Pengu14 wrote:

e_is_2.71828 wrote:

Pengu14 wrote:

Both solutions are wrong.

Oshawoot wrote:

S1973

Actually, your claim that the solutions are wrong is wrong. The issue is that in claiming $n$ to be the greatest positive integer, and so $n^2 \leq n$ , you are contradicting the proposition I proved, which is why your solution is wrong. In fact, an easy proof that $\mathbb{N}$ is unbounded is by letting $f(n)=n^2$ , and $f(n_i) \in \mathbb{N}$ for all $n_i \in \mathbb{N}$ , and $n_i < f(n_i)$ eventually for large enough $n_i$ (technically just greater than 1 lol), so $\mathbb{N}$ is thus unbounded. So both proofs are actually correct, and Oshawoot's solution is the incorrect one, as it fails to prove why the set of natural numbers is not bounded.

Did bro even read my “proof” :rotfl:

What's hilarious is that you think you did something. I wrote what I had to say, if you simply can't understand that a "proof" in which you assume a statement $p$ but $p$ can be shown to be false then your proof is invalid. It's obvious that your assumption logically entailed something false.

Given:
If $p$ , then $q$ .
Not $q$ .
Therefore, not $p$ .

If there is a largest natural number $n$ , then $n^2$ being a distinct natural number must be smaller.
It is not true that $n^2 < n$ .
Therefore, there is no largest natural number. $\blacksquare$

This post has been edited 1 time. Last edited by e_is_2.71828, Apr 21, 2025, 12:42 PM
Reason: Edit

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e_is_2.71828

#8177 h Apr 21, 2025, 12:42 PM

Y by

iwastedmyusername wrote:

ok admittedly i just wanted to use the "your claim that the solutions are wrong is wrong is wrong."

I think the real issue with your solution is that you fail to address the root of the bogus proof, that there is no largest real number (which can easily be proven by n+1>n for a supposed largest real number n). If I said "This proof is bogus because 2, a real number, is greater than 1", would that be a valid solution?

I do think so, yes, but I mean it wouldn't address why $2>1$ contradicts your logic/premises somewhere. I pointed directly where the proof doesn't work, and I provided a proof as you said as to why the set of natural numbers is indeed unbounded. I can't see what else I was supposed to do.

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Soupboy0

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Soupboy0

#8178 h Apr 21, 2025, 2:46 PM

Y by

P1975

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Twinotter

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Twinotter

#8179 h Apr 21, 2025, 3:21 PM

Y by

Soupboy0 wrote:

P1975

S1975

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iwastedmyusername

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iwastedmyusername

#8180 h Apr 21, 2025, 3:26 PM

Y by

e_is_2.71828 wrote:

iwastedmyusername wrote:

ok admittedly i just wanted to use the "your claim that the solutions are wrong is wrong is wrong."

I think the real issue with your solution is that you fail to address the root of the bogus proof, that there is no largest real number (which can easily be proven by n+1>n for a supposed largest real number n). If I said "This proof is bogus because 2, a real number, is greater than 1", would that be a valid solution?

I do think so, yes, but I mean it wouldn't address why $2>1$ contradicts your logic/premises somewhere. I pointed directly where the proof doesn't work, and I provided a proof as you said as to why the set of natural numbers is indeed unbounded. I can't see what else I was supposed to do.

As it says in the literal first post, the next person will find the error that the previous person made, not just point out anything that's wrong in the bogus proof. The error isn't that n^2<=n, it's the assumption that there is a largest real number.

This post has been edited 1 time. Last edited by iwastedmyusername, Apr 21, 2025, 3:27 PM
Reason: a

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e_is_2.71828

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e_is_2.71828

#8181 h Apr 21, 2025, 4:17 PM

Y by

Soupboy0 wrote:

P1975

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e_is_2.71828

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e_is_2.71828

#8182 h Apr 21, 2025, 4:18 PM • 1 Y

Y by Exponent11

iwastedmyusername wrote:

e_is_2.71828 wrote:

iwastedmyusername wrote:

ok admittedly i just wanted to use the "your claim that the solutions are wrong is wrong is wrong."

I think the real issue with your solution is that you fail to address the root of the bogus proof, that there is no largest real number (which can easily be proven by n+1>n for a supposed largest real number n). If I said "This proof is bogus because 2, a real number, is greater than 1", would that be a valid solution?

I do think so, yes, but I mean it wouldn't address why $2>1$ contradicts your logic/premises somewhere. I pointed directly where the proof doesn't work, and I provided a proof as you said as to why the set of natural numbers is indeed unbounded. I can't see what else I was supposed to do.

As it says in the literal first post, the next person will find the error that the previous person made, not just point out anything that's wrong in the bogus proof. The error isn't that n^2<=n, it's the assumption that there is a largest real number.

An error is what the person did wrong in the previous proof...

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Pengu14

577 posts

Pengu14

#8183 h Apr 21, 2025, 10:28 PM

Y by

e_is_2.71828 wrote:

If $n \in \mathbb{N}_*,$ then $n^2 \geq n$ , not the other way around. This is true as $n \geq 1$ and so multiplying by $n$ ( $n$ being positive) we have $n \cdot n \geq n \iff n^2 \geq n$ .

The fact that n^2>n for n>1 is the reason for the contradiction. If n>1 is the largest real number, then all other real numbers are less than or equal to it (including n^2), which contradicts the fact that n^2>n for n>1. The actual mistake (which you failed to mention) was the incorrect assumption that a largest real number exists.

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e_is_2.71828

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e_is_2.71828

#8185 h Apr 22, 2025, 12:06 PM

Y by

Pengu14 wrote:

e_is_2.71828 wrote:

If $n \in \mathbb{N}_*,$ then $n^2 \geq n$ , not the other way around. This is true as $n \geq 1$ and so multiplying by $n$ ( $n$ being positive) we have $n \cdot n \geq n \iff n^2 \geq n$ .

The fact that n^2>n for n>1 is the reason for the contradiction. If n>1 is the largest real number, then all other real numbers are less than or equal to it (including n^2), which contradicts the fact that n^2>n for n>1. The actual mistake (which you failed to mention) was the incorrect assumption that a largest real number exists.

Think of it like this:

Towards a contradiction (which you didn't write because you sought to prove it), assume $\mathbb{N}$ is bounded above by some number $k$ . Then, $\mathbb{N}$ has a supremum, call it $n$ , and $n \in \mathbb{N}$ . Then, $n^2 \leq n$ , but we know $n^2 \geq n$ , so $n=n^2$ . For $n \in \mathbb{N}$ , $n^2=n \iff n=0$ or $n=1$ . Regardless, $2>1>0$ , which is a contradiction since $2 \in \mathbb{N}$ , thus the original assupmtion that $\mathbb{N}$ is bounded is wrong by Reductio Ad Absurdum.

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