# Need help on these proof questions

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Questions: http://prntscr.com/rmuy7e

For Q11 c) I think there is a typo in the question as I can only prove it if it's - a + b - c + d:

n = 1000a + 100b + 10c + d

n = 1001a - a + 99b + b + 11c - c + d

n = 11(91a + 9b + c) - a + b - c + d

For Q11 a) and Q12, I'm not sure it's even possible to 'prove'. In the textbook it just says 'position value' for Q11a and 'consider the rationality of sqrt(2)^sqrt(2)' for Q12.

For Q11 c) I think there is a typo in the question as I can only prove it if it's - a + b - c + d:

n = 1000a + 100b + 10c + d

n = 1001a - a + 99b + b + 11c - c + d

n = 11(91a + 9b + c) - a + b - c + d

For Q11 a) and Q12, I'm not sure it's even possible to 'prove'. In the textbook it just says 'position value' for Q11a and 'consider the rationality of sqrt(2)^sqrt(2)' for Q12.

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Questions: http://prntscr.com/rmuy7e

For Q11 c) I think there is a typo in the question as I can only prove it if it's - a + b - c + d:

n = 1000a + 100b + 10c + d

n = 1001a - a + 99b + b + 11c - c + d

n = 11(91a + 9b + c) - a + b - c + d

For Q11 a) and Q12, I'm not sure it's even possible to 'prove'. In the textbook it just says 'position value' for Q11a and 'consider the rationality of sqrt(2)^sqrt(2)' for Q12.

**TSR360**)Questions: http://prntscr.com/rmuy7e

For Q11 c) I think there is a typo in the question as I can only prove it if it's - a + b - c + d:

n = 1000a + 100b + 10c + d

n = 1001a - a + 99b + b + 11c - c + d

n = 11(91a + 9b + c) - a + b - c + d

For Q11 a) and Q12, I'm not sure it's even possible to 'prove'. In the textbook it just says 'position value' for Q11a and 'consider the rationality of sqrt(2)^sqrt(2)' for Q12.

Q12) the example shows that irrational^irrational can be rational. It can also be irrational. Obviously the example given is rational. If sqrt(2)^sqrt(2) is rational, result is proved. If it's irrational, you've just shown when you raise it to sqrt(2) it's rational.

Not really sure what your problem is with Q11a)?

Last edited by mqb2766; 1 year ago

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(Original post by

Q11c) Both are correct as -121 is divisible by 11, just as is 121. You could get the book answer by flipping the 1001 or 999 in the original subtraction.

Q12) the example shows that irrational^irrational can be rational. It can also be irrational. Obviously the example given is rational. If sqrt(2)^sqrt(2) is rational, result is proved. If it's irrational, you've just shown when you raise it to sqrt(2) it's rational.

Not really sure what your problem is with Q11a)?

**mqb2766**)Q11c) Both are correct as -121 is divisible by 11, just as is 121. You could get the book answer by flipping the 1001 or 999 in the original subtraction.

Q12) the example shows that irrational^irrational can be rational. It can also be irrational. Obviously the example given is rational. If sqrt(2)^sqrt(2) is rational, result is proved. If it's irrational, you've just shown when you raise it to sqrt(2) it's rational.

Not really sure what your problem is with Q11a)?

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I don't think Q11a is possible to prove. The best 'proof' I can come up with is proof by example, but in the textbook it says Position value. How do I prove a position value?

**TSR360**)I don't think Q11a is possible to prove. The best 'proof' I can come up with is proof by example, but in the textbook it says Position value. How do I prove a position value?

23 = 20+3 = 2*10 + 3

I think that's all that's required.

Last edited by mqb2766; 1 year ago

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