Proofs help (Proof by Contradiction) (1 Viewer)
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1. cos(n+1) = 2cos(n)cos(1) - cos(n-1)
Need help with part b) from the first question and the other 2 questions.
Suppose cos 1 is rational.
It follows that cos 1,2,3,4,5.... is rational.
But, there exists cos n such that cos n is irrational, e.g., cos 30.
2. can't work out the question. If a is not a factor of b, it is obviously not factor of b+1 if a>= 2.
3. y^2 = x^2 -1
y^2 = (x+1)(x-1)
by definition, a square number is made up of the product of the same numbers.
if b is an integer, it must be a factor of +- itself, b or -b. If it is not a factor of any natural number, it is an integer that is not a natural number and is thus 0.
this question seems like nonsense tho, maybe im understanding it wrong.
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a) is just rational root theorem. Oh you want by contradicton. Assume
b) Assume a and b are integers. The expression is equivalent to:
1. Prove that there is no integer that leaves a remainder of 2 on division by 6, and also leaves a remainder of 7 on division by 9.
2. Prove that
for any value 
3. Prove that no integer has a square that leaves a remainder of 3 on division by 4.
4. Prove that if a right angled triangle has sides of lengths
, 
, and 
(where is the length of the hypoteneuse), and another right angled triangle has sides of lengths 
, 
, and 
(where is the length of the hypoteneuse), then at least one of , , and is not an integer.
5. Prove that
is irrational.
6(a). Suppose that and are positive integers. Find real constants 
and 
such that
(a+b+1)} = \frac{A}{a+b} + \frac{B}{a+b+1})
(b). Hence, or otherwise, show that, if
is a positive integer, then
(n+k+1)}}<\frac{1}{n+1})
(c). You may use, without proof, the fact that
in order to prove that
is irrational.
2. Prove that
3. Prove that no integer has a square that leaves a remainder of 3 on division by 4.
4. Prove that if a right angled triangle has sides of lengths
5. Prove that
6(a). Suppose that
(b). Hence, or otherwise, show that, if
in order to prove that
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Qeru, just a quick reminder that you needa) is just rational root theorem. Oh you want by contradicton. Assume(where a and b are non zero integers) is a root. Then:
. The parity of the LHS is different to the parity of the RHS so contradiction.
As it happens, this is necessary for the parity argument that you are making in this case. You are stating that the LHS is odd and the RHS is even, which is true but only because at least one of
Further, examiners will not necessarily know what you mean by LHS and RHS have different parity. It would be simpler in this case to note
How does this follow? You could have some combination of a and b which equals zero (since either a or b could be negative).Further, examiners will not necessarily know what you mean by LHS and RHS have different parity. It would be simpler in this case to noteand so
, producing the necessary contradiction.
True... an alternative is then that
How could we do Question 4)1. Prove that there is no integer that leaves a remainder of 2 on division by 6, and also leaves a remainder of 7 on division by 9.
2. Prove that
3. Prove that no integer has a square that leaves a remainder of 3 on division by 4.
4. Prove that if a right angled triangle has sides of lengths
5. Prove that
6(a). Suppose that
(b). Hence, or otherwise, show that, if
(c). You may use, without proof, the fact that
in order to prove that
Assume a,b and c are integers. We know from pythagoras:
But addition is closed for the integers so the LHS must be an integer but the RHS is clearly not so contradiction. So atleast one of a,b or c is not an integer.