Question

If (2x-5)/(x-4) = 3, then x = 7
What is wrong with this algebraic proof?
suppose x = 7, then (2*7-5)/(7-4) = 3. Therefor,
if (2x-5)/(x-4) = 3, then x = 7.

Answer 1

I know you can't assume the conclusion, but isn't what we do to check our solutions?

Answer 2

thats trial and error method

Answer 3

yes, thats how we make sure that x=7 is a solution.

Answer 4

I agree with @hartnn

Answer 5

The thing that is wrong with this proof is that it uses the converse of a statement, which is not true in all cases. Let me give you an example:

If \(x = 3\), \(x^2 - 10x + 21 = 0\). This one is true.

However,

If \(x^2 - 10x + 21 = 0\), \(x = 3\)

is not true at all.

Answer 6

its partly true

x=3 or x=7 is completely true

Answer 7

Hehe, but it's a false statement, as there are only true and false here.

Answer 8

i would have gone with true, because its an "or" there
and not "and"

Answer 9

If x2−10x+21=0, x=3 TRUE

Answer 10

No, not talking about your statement. I'm talking about mine... what I meant was "partly true" isn't a thing.

Answer 11

huhm... a statement can't be partially true though

Answer 12

what i was saying is its not false, for sure.

Answer 13

I guess it is. If it's not true, then it is false.

Answer 14

why not ?
there can be incomplete information, and that info. can be true.

Answer 15

The equation (2x-5)/(x-4) = 3 is linear.
Hence, your method is a valid proof.

But if the equation wasn't linear, then your solution would have been called "incomplete"

Answer 16

I mean a counterexample to that statement is \(x = 7\). If there is a counterexample, then the statement is false.

Answer 17

@LastDayWork perhaps you meant one-to-one? (2x-5)/(x-4) certainly isn't linear

Answer 18

its not counter example

if x^2 =1, x =1 true
if x^2 = 1, x= -1 true
if x^2=1, x =1 or x=-1 true
if x^2 =1 , x=0 false

Answer 19

@ParthKohli
both are true in your example
you did not set a parameter that both x's be satisfied

Answer 20

I mean it can be reduced to linear, or one-one..

Answer 21

(2x-5)/(x-4) = 3 is not linear
(2x-5)=3(x-4) is linear.

Answer 22

^^

Answer 23

OK -- let me rephrase that

If x equals three, then x^2 - 10x + 21 must equal 0.

and

If x^2 - 10x + 21 equals zero, then x must equal three.

Answer 24

lol true again


x = 3 ONLY ----> False

Answer 25

No... look... "x must equal 3" is wrong. x can equal 7.

Answer 26

but you forgot to set iff and only iff

Answer 27

x can =7 does not mean x must not be 3.

Answer 28

wholeheartedly agrees with @hartnn the only way it will be a false statement is if you strictly made it only x=3 to satisfy the equation

Answer 29

^^
x= 3 only is false, as already told.

Answer 30

By "must equal three", I did mean "three ONLY"

Answer 31

The general statement;
Suppose x = a ; then f(x) = 0
Therefore, if f(x) = 0 then x = a

is wrong as there can be a number b such that;
f(b) = 0 but b≠a

I guess, that's what @ParthKohli implies..

Answer 32

you should have specified 'only', else it changes things.

Answer 33

"MUST equal three"

Answer 34

Suppose x = a ; then f(x) = 0
Therefore, if f(x) = 0 then x = a
TRUE

Answer 35

Er..
P => Q
is valid only iff whenever P is true, Q is also true

Answer 36

The problem is like @ParthKohli already said. Proving the converse doesn't prove the original statement.

Answer 37

"MUST" is the same as "only", more or less. It's a word that restricts.

Answer 38

x^2−10x+21= 0 is a true statement if the conclusion is x = 3 or x = 7

x2−10x+21=0, x=3 False
x2−10x+21=0, x=7 false
x2−10x+21=0, x=3 or x = 7 True

Answer 39

all fat cows are female
therefore, all females are fat cows
a ⇒ b
b ⇏ a

iff you're an egomaniacal douchebag
a ⇔ b

Answer 40

The contrapositive is what you would need to use. In this case if \(x\neq 7\) then \((2x5)/(x-4)\neq -3\)

Answer 41

Or, in terms of Venn diagram
solution set for P should be completely inside the solution set for Q.

Hence,
Suppose x = a ; then f(x) = 0
Therefore, if f(x) = 0 then x = a

is False

Answer 42

thank you for everyone's time =]