Arjun asked in Science & MathematicsMathematics · 9 years ago

# Mathematics induction question?

Prove by mathematical induction that n < 2n for all positive integers n

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• Brenda
Lv 7
9 years ago

n<2n

1)check for n=1

1<2*1

1<2 true

2)n=k

k<2k

3)n=k=k+1

k+1<2(k+1)

k+1<2k+2

remember k<2k by induction assumption

and of course 1<2 is always true

thus

k+1<2k+2 is true for all positive integers n

n<2n true

Lv 7
9 years ago

1 < 2·1, so the premise holds true for n = 1.

Suppose the premise is true for some value of n that is greater than 1. Then

n < 2n

Add 1 to both sides of the inequality.

n + 1 < 2n + 1

Since n is positive,

2n + 1 < 2n + 2

2n + 2 = 2(n + 1)

By transitivity,

n + 1 < 2(n + 1)

Therefore, the premise holds true for the case n + 1 as well.

• 9 years ago

property is valid for n = 1

1 < 2

suppose property holds for n

n < 2n

we must prove that it is true for n + 1, that is

n + 1 < 2(n + 1)

n + 1 < 2n + 2

n < 2n + 1

since n < 2n to a greater extent n < 2n + 1

in other words

if n < 2n < 2n + 1

for transitive property of < we have

n < 2n + 1

QED

• Luis
Lv 6
9 years ago

Dear Arjun,

n<2n

Since 2n contains the variable to solve for, move it to the left-hand side of the inequality by subtracting 2n from both sides.

n-2n<0

According to the distributive property, for any numbers a, b, and c, a(b+c)=ab+ac and (b+c)a=ba+ca. Here, n is a factor of both n and -2n.

(1-2)n<0

To add integers with different signs, subtract their absolute values and give the result the same sign as the integer with the greater absolute value. In this example, subtract the absolute values of 1 and -2 and give the result the same sign as the integer with the greater absolute value.

(-1)n<0

Remove the parentheses.

-n<0

Multiply each term in the inequality by -1. When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

-n*-1>0*-1

Multiply -n by -1 to get n.

n>0*-1

Multiply 0 by -1 to get 0.

n>0

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Thanks

Source(s): Precalculus Solved!