Question

check whether π - 22/7 a rarional number. justify your answer.

Answer 1

No π - 22/7 is not a rational number, this can be explained as follows,

We know rational number can be put in form p/q where p & q are rational numbers and q is not equal to 0

But, π on the contrary is a rational number because it's non-recurring and non-repeating and can't be put on the number line.

Also when you subtract a rational number from an irrational number it always results in an irrational number

Hence, π - 22/7 is irrational.

Should I give mathematical proof also?

Mathematical Proof:

Suppose instead that π is rational. Then there exists integers a and b with b≠0 such that π=ab. Define:

(1)

f(x)=xn(a−bx)nn!

(2)

F(x)=f(x)−f(2)(x)+f(4)(x)−...(−1)nf(2n)(x)

Note that f(k)(0) and f(k)(π) are integers for every nonnegative integer k.

We now compute the derivative of F′(x)sinx−F(x)cosx by the product rule for differentiation to get:

(3)

[F′(x)sinx−F(x)cosx]′=F′′(x)sinx+F′(x)cosx−F′(x)cosx+F(x)sinx=F′′(x)sinx+F(x)sinx=[F′′(x)+F(x)]sinx

Observe that:

(4)

F′′(x)+F(x)=[f(x)−f(2)(x)+f(4)(x)−...(−1)nf(2n)(x)]′′+[f(x)−f(2)(x)+f(4)(x)−...(−1)nf(2n)(x)]=[f(2)(x)−f(4)(x)+...+(−1)nf(2n+2)(x)]+[f(x)−f(2)(x)+f(4)(x)−...(−1)nf(2n)(x)]=f(x)+(−1)nf(2n+2)(x)

But the (2n+2)th derivative of f is zero since f(x) is a polynomial of degree 2n. So F′′(x)+F(x)=f(x) and hence:

(5)

[F′(x)sinx−F(x)cosx]′=f(x)sinx

We integrate f(x)sinx over [0,π] by using the Fundamental theorem of Calculus to get:

(6)

∫π0f(x)sinxdx=[F′(x)sinx−F(x)cosx]π0=[F′(π)sinπ−F(π)cosπ]−[F′(0)sin0−F(0)cos0]=F(π)+F(0)(∗)

The values of F(π) and F(0) are completely determined by sums of the values of f(k)(0) and f(k)(π) which we have already noted to be integer values. So F(π)+F(0) is an integer.

We now show that F(π)+F(0) is a positive integer. Observe that on the interval (0,π) we have that f(x)>0 and sinx>0. So f(x)sinx>0 on this interval. So from (∗) we must have that F(π)+F(0)>0.

Now note that for every x∈(0,π) we have that 0<sinx<1. So on this interval, 0<f(x)sinx<f(x). Also on this interval, a−bx<a. So (a−bx)n<an. But also xn<πn on this interval. So f(x)<πnann!. Putting this all together tells us that for every x∈(0,π) we have that:

(7)

0<f(x)sinx<πnann!

Integrating both sides of this equation on [0,π] gives us:

(8)

0<F(π)+F(0)=(∗)∫π0f(x)sinxdx<∫π0πnann!dx=πn+1ann!

But as n→∞, the righthand side of the inequality above goes to 0 - which is a contradiction, since F(π)+F(0) is supposed to be a positive integer. Therefore the assumption that π was rational was false. So π is irrational. ■

Hence pi is irrational

Answer 2

substitute value of a in equation 1

2a + b = -6

2(-3) + b = -6

-6 +b = -6

b = (-6)+(+6)

b = (-6+6)

b = 0
this is the answer is question:
if the zeroes are 2 and...