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DaniilM [7]
3 years ago
13

How, Let a = a rational number. Let b = an irrational number. Assume a + b = c and c is rational (attempting to disprove the con

jecture). a + b = c b = c - a Explain how Alfred's argument contradicts his initial assumption proving that the sum cannot be rational?
Mathematics
2 answers:
alisha [4.7K]3 years ago
7 0
If a and c are rational then a can be written as x/y and c  can be w/z  where w,x,y and z are integers

so as alfred said  if  b = c - a = x/y - w/z  =  (zx - wy) / yz  which is  rational.
So b cannot be irrational..
laiz [17]3 years ago
6 0
If 'a' is a rational number and c is rational, then 
a = p/q
c = r/s
where p,q,r,s are integers (q and s can't be zero)

Subtracting c-a gives
b = c-a
b = (p/q) - (r/s)
b = (ps/qs) - (qr/qs)
b = (ps - qr)/(qs)

The quantity pq - qr is an integer. The reason why is because ps and qr are both integers (multiplying any two integers leads to another integer). Subtracting any two integers results in another integer.

So we have (ps - qr)/(qs) in the form (integer)/(integer) = rational number

Therefore, b is a rational number, but this contradicts the given info that b is irrational. If b is irrational, then we CANNOT write it as a ratio of integers.

This contradiction proves the assumption "a+b = c and c is rational" is incorrect
The sum is irrational.

Therefore, if a+b = c, where 'a' is rational and b is irrational, then c is irrational.
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  • A) k = 18.75
  • B) R(0.8) = 161419
  • C) R(0.3) = 275

Step-by-step explanation:

<u>Given expression</u>

  • R(x) = 6*e^(kx)

A) <u>Given</u>

  • x = 0.04
  • R = 10

<u>Solving for k</u>

  • 10 = 6*e^(0.04k)
  • e^(0.04k) = 10/6
  • ln (e^0.04k) = ln (1.6666)
  • 0.04k = 0.51
  • k = 0.51/0.04
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B) <u>Given</u>

  • x = 0.8
  • R= ?

<u>The value of R(0.8) is:</u>

  • R(0.8) =
  • 6*e^(0.8*12.75) =
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C) <u>Given</u>

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  • R = ?

<u>The value of R(0.3) is:</u>

  • R(0.3) =
  • 6*e^(0.3*12.75) =
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