In the class we've done a bit of induction but one of my homework questions was this and I'm completely lost, any suggestions please?
Show that 2 −√2 is irrational. (Hint: Assume that √2 is irrational, and use ‘Proof by
Contradiction’.
Show that 2 −√2 is irrational. (Hint: Assume that √2 is irrational, and use ‘Proof by
Contradiction’.
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OK, proof by contradiction starts by assuming the opposite.
So suppose that 2 - √2 is rational, that means that it can be represented by a fraction, such as b/a.
You have
2 - √2 = b/a
Multiply by a :
2a - a√2 = b
Rearrange :
a√2 = 2a - b
√2 = (2a - b)/a
(2a - b)/a is a rational number ; but you are told to assume that √2 is irrational. Therefore you have a rational number equal to an irrational number. Which is a contradiction. Therefore your starting assumption is wrong : 2 - √2 is not rational.
So suppose that 2 - √2 is rational, that means that it can be represented by a fraction, such as b/a.
You have
2 - √2 = b/a
Multiply by a :
2a - a√2 = b
Rearrange :
a√2 = 2a - b
√2 = (2a - b)/a
(2a - b)/a is a rational number ; but you are told to assume that √2 is irrational. Therefore you have a rational number equal to an irrational number. Which is a contradiction. Therefore your starting assumption is wrong : 2 - √2 is not rational.
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Proof by contradiction is always set up this way:
Prove
Assume
Show by a sequence of steps how this assumption leads to a contradiction like in the previous post, show how a/b has no common factors but then given the assumption, it HAS TO HAVE common factors since both are even.
When done properly, this is all that is necessary to Prove the assertion is true.
These proofs are often short and are very powerful. However, some consider them a little soft, lacking rigor, as a means for developing theorems.
But, when in a pinch with a proof, proof by the method of contradiction is very powerful.
Prove
Assume
Show by a sequence of steps how this assumption leads to a contradiction like in the previous post, show how a/b has no common factors but then given the assumption, it HAS TO HAVE common factors since both are even
When done properly, this is all that is necessary to Prove the assertion is true.
These proofs are often short and are very powerful. However, some consider them a little soft, lacking rigor, as a means for developing theorems.
But, when in a pinch with a proof, proof by the method of contradiction is very powerful.
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If √2 were rational number let √2 = a/b where a and b have no common factors (a and b are coprimes)
So a^2 = 2b^2 which means a^2 and hence a is an even number. Say a = 2m. So a^2 = 4m^2
but then 4m^2 = 2b^2 means b^2 = 2m^2 and b is also even.
Which means a and b have common factor #(a contradiction). Hence he assumption must be wrong
Cleary 2−√2 is irrational.
FYI, this is an elemntary proof and has nothing to do with induction]
So a^2 = 2b^2 which means a^2 and hence a is an even number. Say a = 2m. So a^2 = 4m^2
but then 4m^2 = 2b^2 means b^2 = 2m^2 and b is also even.
Which means a and b have common factor #(a contradiction). Hence he assumption must be wrong
Cleary 2−√2 is irrational.
FYI, this is an elemntary proof and has nothing to do with induction]