How can 1 = 0 ?
Peter has enough! All these mathematics do not make any sense to him and he is ready to prove to his teacher that mathematics is based on illogical principles!
The way he intends to do this, is by proving that 1=0. Here is his argument :
{1} Suppose that
x
satisfies :
x
−
1
=
0
.
{2} But then that
⟹
(
x
−
1
)
(
x
−
2
)
=
0
{3} which implies
x
−
1
=
0
or
x
−
2
=
0
{4} which implies
x
=
1
or
x
=
2
{5} So, in particular,
x
=
2
is a solution to the first equation and hence, by substituting in, we obtain :
1
=
0
Can you tell poor Peter where his argument goes wrong?
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2 solutions
statement 3 also say the same thing as statement 5 : writing x - 2 = 0 is incorrect.
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The either/or construct generally has as a minimum requirement that at least one of the elements is true. The inclusive or allows for the possibility that both can be true but does not require this. (The exclusive or does not allow for both to be true; only one or the other, but not both.) So technically both statements {3} and {4} are valid. Also, in isolation of supposition {1}, statement {3} does follow from statement {2}, and {4} from {3}. Everything comes crashing down though, logically speaking, with statement {5}. where the either/or construct of {3} and {4} has been violated. The construction of the problem is deceptive because we know that x − 2 = 0 is 'false bait' from the start, but it is only when this 'false bait' is taken as the real thing in statement {5} that the logical argument truly breaks down.
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Thanks understood my mistake , so writing x - 2 =0 is incorrect ,due to the word "OR" we are just expressing the possibility. Agreed 5 is incorrect
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@U Z – Exactly. I nearly chose {3} when I first came across this question until I started looking at it from a purely technical perspective. It's a good question; it made me think. :)
@U Z – @megh choksi I just had a look at your problem "How many methods for this?" and have a question to ask. Is the blue line supposed to be a diameter of the green circle, or are the two circles supposed to be concentric? As it stands the green circle can be any circle with a diameter ≥ 7 4 , so we need to know some other defining feature to make it unique.
@U Z – i also think that 2 and 3 are wrong as if we start multiplying zero with anything then we can prove almost any thing. the 5th part is fine
Ya the wordings of question should be changed
True. I am thinking of making a small edit to the statement of the problem though. Do you think that in statement number 5, we should write 'solution to first equation' instead of 'solution to above equation' because there are two equations?
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That sounds like a good idea; I interpreted it as referring to the original equation but it might help to make it more precise.
The actual mistake is that the implications are only in one direction, which is why the solutions obtained in {5} are not necessarily the solutions to {1}. (A priori, we do not even know if at least one of them is the root.
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I suppose that, since the implications are in only one direction, that if neither x = 1 nor x = 2 were solutions then there would be no solution to the original equation. But if we can state that the original equation has a solution, (not worrying about what it actually might be), then it must be either x = 1 or x = 2 , (or both). So the mistake, (or at least one of them), in {5} is to conclude that x = 2 must be a solution, an assertion that isn't supported by the previous statements.
I was thinking of how this might play out if it was not at all obvious that the original equation we chose had any solutions, and we just supposed it did. Could that line of thought ever be useful, or would we just be spinning our wheels? I guess we could end up proving that there is no solution if we come up with an absurdity, but I'm just wondering if it could shed some light on any potential solutions, if not actually find them outright.
And also by supposition... only x is variable .... therefore there is only one solution not two solutions.
I consider the third relation as wrong. Because (x-1)=0 is multiplied by (x-2). Then we got (x-1)(x-2)=0. But this is a wrong process. As an example if i multiply the equation (x-1)=0 with 5 then the equation is (x-1)*5=0 so we said from that x-1=0 or 5=0 so we can prove every thing from here. So i think the 3 is wrong
By multiplying by (x-2) in stage 2, we assume that (x-2) is different from 0 (multiplying by 0 is 'illegal' in equations) and thus we can't conclude that x=2 is a solution in 5.
But we know from the supposition that x − 1 = 0 , so any product involving x − 1 and any finite quantity, including 0 , will yield a result of 0 . Thus we don't actually have to make any assumptions about x − 2 when we introduce it in stage 2 .
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But if at any stage you multiply both sides by (x-a), you're immediately canceling the option of x=a that's basic algebra...
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But if x − 1 = 0 , then ( x − 1 ) ∗ A = 0 regardless of what A is, (as long as it's finite).. In this case we happen to have A = x − 2 , and as A can have any finite value, including 0 , we aren't having to rule out x = 2 because we are multiplying by x − 2 , but because we are basing the validity of the equation ( x − 1 ) ( x − 2 ) = 0 on the supposition that x − 1 = 0 . The introduction of the factor ( x − 2 ) is a bit of a con game, which Peter tries to play out in step 5, which is why this is the step where the logic of his argument fails.
Correction: If you divide both sides of an equation by ( x − a ) , then you must have x = a . Multiplication doesn't affect the equation as long as the value which you're multiplying on both sides is finite (defined).
Looking at the steps one at a time ......
{1} The given equation has a solution, so it is valid to make this supposition.
{2} By the zero product property, since x − 1 = 0 we know that ( x − 1 ) ( x − 2 ) = 0 is a valid equation.
{3} Since x − 1 = 0 , this either/or statement includes a correct option and hence is a valid statement.
{4} Since x − 1 = 0 ⟹ x = 1 , this either/or statement includes a correct option and hence is a valid statement.
{5} It is not true that from statement {4} we can conclude that both x = 1 and x = 2 are solutions, only that at least one of them is. So this is where Peter's logic has failed. It happens to be the case that only one of the options is in fact a solution, namely x = 1 .