# Irrationality of this number. Proof.

## Homework Statement

Prove that if x is a real number then 2x2-3=0

## The Attempt at a Solution

Well I think this proof is OK but it seems very to the point, please tell me if there is any fallacy in my reasoning.

2x2-3=0

2x2=3

but 2x2 is even, while 3 is odd

Hence x must be irrational for all real x.

??

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Prove that if x is a real number then 2x2-3=0

## The Attempt at a Solution

Well I think this proof is OK but it seems very to the point, please tell me if there is any fallacy in my reasoning.

2x2-3=0

2x2=3

but 2x2 is even, while 3 is odd

Hence x must be irrational for all real x.

??
All you proved is that x in not an integer.

What is the problem as stated for you to solve?

Hurkyl
Staff Emeritus
Gold Member
There are many things wrong. if x is a real number then 2x2-3=0
This statement is false. x=0 is a counterexample: it satisfies the hypothesis, but not the conclusion.

2x2-3=0

2x2=3
For this bit, I can't tell if there's any problem with your reasoning, because you don't say what you're reasoning is -- you just wrote two equations.

but 2x2 is even, while 3 is odd
The notion of "even" and "odd" applies to integers, but it doesn't apply to real numbers. If you are using x to denote a real number, then 2x2 is also 'merely' a real number, and so it doesn't make sense to talk about it being even.

Hence x must be irrational for all real x.
This is another false statement. x=0 is another counterexample. It is real, however it is not irrational. It also has nothing to do with the statement you said you were trying to prove.

I hate to say it, but I think what you need to do is to focus on trying to understand logic, grammar, and the meaning of what you read/write, and also on explicitly writing out what you are thinking and doing.

Would you please state exactly the question you were asked? I'm guessing it's something similar to
If x is a real number satisfying the equation 2x2 - 3 = 0, then x is irrational.​
Once you've stated the question, it may be useful to do one or all of the following:
• Explain why your statement in the opening post means the same thing or doesn't mean the same thing as the question you were asked.
• Explain why my statement means the same thing or doesn't mean the same thing as the question you were asked.
• Explain why my statement means the same thing or doesn't mean the same thing as your statement in the opening post.

There are many things wrong. This statement is false. x=0 is a counterexample: it satisfies the hypothesis, but not the conclusion.

For this bit, I can't tell if there's any problem with your reasoning, because you don't say what you're reasoning is -- you just wrote two equations.

The notion of "even" and "odd" applies to integers, but it doesn't apply to real numbers. If you are using x to denote a real number, then 2x2 is also 'merely' a real number, and so it doesn't make sense to talk about it being even.

This is another false statement. x=0 is another counterexample. It is real, however it is not irrational. It also has nothing to do with the statement you said you were trying to prove.

I hate to say it, but I think what you need to do is to focus on trying to understand logic, grammar, and the meaning of what you read/write, and also on explicitly writing out what you are thinking and doing.

Would you please state exactly the question you were asked? I'm guessing it's something similar to
If x is a real number satisfying the equation 2x2 - 3 = 0, then x is irrational.​
Once you've stated the question, it may be useful to do one or all of the following:
• Explain why your statement in the opening post means the same thing or doesn't mean the same thing as the question you were asked.
• Explain why my statement means the same thing or doesn't mean the same thing as the question you were asked.
• Explain why my statement means the same thing or doesn't mean the same thing as your statement in the opening post.

Ok I miss read the question and I thought it was for x an integer...

is this better.

Suppose that

2x^2-3=0

adding 3 two both sides of the equation

2x^2=3

dividing both sides by 2

x^2=3/2

taking the square root of both sides

x=√(3/2)

Now suppose that x is not irrational, ie suppose

x=p/q for some integers p/q

so √(3/2)=p/q

so √3/√2 =p/q

but √3 and √2 are both irrational.

Contradiction therefore x must be irrational.

Office_Shredder
Staff Emeritus
Gold Member
The ratio of two irrational numbers can be rational, for example the square root of eight divided by the square root of two.

Do you know how to prove that the square root of two is irrational? A similar argument can be used here

The ratio of two irrational numbers can be rational, for example the square root of eight divided by the square root of two.

Do you know how to prove that the square root of two is irrational? A similar argument can be used here

Yeah I think it should be along the lines...

Assume that √2 is not irrational ie

√2=a/b (by definition of rationality).

so 2=a^2/b^2

so..

2b^2=a^2

Now a^2 is divisible by 2 and clearly so is a.

so let a=2k

so 2b^2=(2k)^2

2b^2=4k^2

b^2=2k^2

so b^2 is also divisible by 2.

Contradiction Since any rational number is a quotient or relative primes.

Therefore √2 is rational.

HallsofIvy
You last statement should be "$\sqrt{2}$ is NOT rational"!
Now apply those same ideas to $\sqrt{3/2}$.
You last statement should be "$\sqrt{2}$ is NOT rational"!
Now apply those same ideas to $\sqrt{3/2}$.