Here's my problem:(adsbygoogle = window.adsbygoogle || []).push({});

Provide either a proof or a counterexample for each of these statements.

a) For all real numbers x and y, if x is greater than 1 and y is greater than zero, then y^x is greater than x.

My proof:

Suppose x is some real number greater than 1 and y is some real number greater than 0.

Suppose x=2 and y=1/4.

Then y^x=(1/4)^2=1/16 and 1/16=y^x is less than x=2.

Now suppose that x=3 and y=2.

Then y^x=2^3=8 and 8=y^x is greater than x=3.

Hence if x is greater than 1 and 0 less than y less than or equal to 1, then y^x is less than or equal to x.

But if x is greater than 1 and y is greater than 1, then y^x is greater than x.

Therefore the statement "if x is greater than 1 and y is greater than 0, then y^x is greater than x" is not true for all real numbers x and y.

Is this a good proof? How can I improve it or make it clearer?

b) For integers a, b, c, if a divides bc, then either a divides b or a divides c.

I'm not really sure where to go with this one, so hints would be welcome.

I do know that if a divides bc, then bc=ak, where k is a natural number.

Similarly, a divides b means that b=aj and a divides c means that c=ai, where j and i are also natural numbers.

Which proof techinique do I use here? contradiction, contraposition, or direct proof?

Thanks ahead of time,

eku_girl83

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# Need help with two simple proofs

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