1) True.
\(\displaystyle{x}^{{2}}-{5}{x}={y}^{{2}}-{5}{y}\)
Rearranging equation,
\(\displaystyle{x}^{{2}}−{y}^{{2}}={5}{x}−{5}{y}\)
\(\displaystyle{x}^{{2}}−{y}^{{2}}={5}{\left({x}−{y}\right)}\)
Using identity \(\displaystyle{x}^{{2}}−{y}^{{2}}={\left({x}+{y}\right)}{\left({x}−{y}\right)}\) we get,
\(\displaystyle{\left({x}+{y}\right)}{\left({x}−{y}\right)}={5}{\left({x}−{y}\right)}\)
Dividing by (x−y) on both sides we get,
x+y=5
Therefore, if \(\displaystyle{x}^{{2}}−{5}{x}={y}^{{2}}−{5}{y}\) and \(\displaystyle{x}\ne{y}\), then x+y is five.
2) False.
The decimal expansion of pi is 3.1415926535897932384....., the decimal does not repeat exactly. pi is an irrational number. Therefore, it cannot be expressed as ratio of two integers and it cannot be expressed as terminating or repeating decimal.
3) True.
Any irrational number divided by an integer is irrational number. Proof (by contradiction):
Let x be a irrational number and a be an integer if \(\displaystyle\frac{{x}}{{a}}\) is a rational number it can be expressed as ratio of two integers, \(\displaystyle\frac{{x}}{{a}}=\frac{{m}}{{n}}\) this implies \(\displaystyle{x}=\frac{{{a}{m}}}{{n}}\). Since a, m and n are integers, am is an integer and x is ratio of two integers which implies x is a rational number. This is a contradiction to fact that x is irrational number.
Therefore, if an irrational number divided by a non zero integer then the result is an irrational number.
\(\displaystyle{x}^{{2}}-{5}{x}={y}^{{2}}-{5}{y}\)
Rearranging equation,
\(\displaystyle{x}^{{2}}−{y}^{{2}}={5}{x}−{5}{y}\)
\(\displaystyle{x}^{{2}}−{y}^{{2}}={5}{\left({x}−{y}\right)}\)
Using identity \(\displaystyle{x}^{{2}}−{y}^{{2}}={\left({x}+{y}\right)}{\left({x}−{y}\right)}\) we get,
\(\displaystyle{\left({x}+{y}\right)}{\left({x}−{y}\right)}={5}{\left({x}−{y}\right)}\)
Dividing by (x−y) on both sides we get,
x+y=5
Therefore, if \(\displaystyle{x}^{{2}}−{5}{x}={y}^{{2}}−{5}{y}\) and \(\displaystyle{x}\ne{y}\), then x+y is five.
2) False.
The decimal expansion of pi is 3.1415926535897932384....., the decimal does not repeat exactly. pi is an irrational number. Therefore, it cannot be expressed as ratio of two integers and it cannot be expressed as terminating or repeating decimal.
3) True.
Any irrational number divided by an integer is irrational number. Proof (by contradiction):
Let x be a irrational number and a be an integer if \(\displaystyle\frac{{x}}{{a}}\) is a rational number it can be expressed as ratio of two integers, \(\displaystyle\frac{{x}}{{a}}=\frac{{m}}{{n}}\) this implies \(\displaystyle{x}=\frac{{{a}{m}}}{{n}}\). Since a, m and n are integers, am is an integer and x is ratio of two integers which implies x is a rational number. This is a contradiction to fact that x is irrational number.
Therefore, if an irrational number divided by a non zero integer then the result is an irrational number.
