Pepto Bismol
New Member
How can you prove that square root 2 is an inifinite number? I tried with p/q=root 2, but it won't work, otherwise I don't understand it.
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Square Root 2 (1 Viewer)
- Thread starter Pepto Bismol
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honky tonk
in Miracle World
Errrrrr.. is this General?
hipsta_jess
Up the mighty red V
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can you please take some of the shit out of your siggy, Pepto Bismol, thanks..its just it takes up a LOT of room and therefore is highly irritating.
also, as honky asked, are you sure this is a general maths question?
and the square root of two is a surd...is that what you mean?
also, as honky asked, are you sure this is a general maths question?
and the square root of two is a surd...is that what you mean?
Dumbarse
Member
GENERAL MATHS
Get with maths people!
Get with maths people!
Dumbarse
Member
Do SOME STUDY
OK FOLKS IT TIME TO WORK!
OK FOLKS IT TIME TO WORK!
honky tonk
in Miracle World
Re: Do SOME STUDY
(This thread is so gonna get deleted.)
But it's such a pretty bottle..
(This thread is so gonna get deleted.)
You don't need to know this for General Maths.
ASSUMPTION: sqrt(2) is rational.
It can therefore be written as the quotient of two integers a and b (where both a and b cannot be even - if they were both even, you could simplify the fraction until either or both turned odd).
sqrt(2) = a/b
2 = (a/b)<sup>2</sup>
2 = a<sup>2</sup>/b<sup>2</sup>
a<sup>2</sup> = 2b<sup>2</sup>
.'. a must be even (twice any number is an even number).
Let a = 2k where k is an integer. Now:
a<sup>2</sup> = 2b<sup>2</sup>
4k<sup>2</sup> = 2b<sup>2</sup>
b<sup>2</sup> = 2k<sup>2</sup>
.'. b must be even (twice any number is an even number).
This is a contradiction, as a and b cannot both be even. The only assumption we made must hence be incorrect.
sqrt(2) is therefore irrational.
ASSUMPTION: sqrt(2) is rational.
It can therefore be written as the quotient of two integers a and b (where both a and b cannot be even - if they were both even, you could simplify the fraction until either or both turned odd).
sqrt(2) = a/b
2 = (a/b)<sup>2</sup>
2 = a<sup>2</sup>/b<sup>2</sup>
a<sup>2</sup> = 2b<sup>2</sup>
.'. a must be even (twice any number is an even number).
Let a = 2k where k is an integer. Now:
a<sup>2</sup> = 2b<sup>2</sup>
4k<sup>2</sup> = 2b<sup>2</sup>
b<sup>2</sup> = 2k<sup>2</sup>
.'. b must be even (twice any number is an even number).
This is a contradiction, as a and b cannot both be even. The only assumption we made must hence be incorrect.
sqrt(2) is therefore irrational.
Gregor Samsa
That Guy
Phew. My abilities at Surds in Year 10 Advanced were minimal. I think I got 9% on the topic test.