WEBVTT
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Hello.
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Let us try to solve the exercise
1 of the rational powers
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in practice step.
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We want to simplify the
following three expressions.
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Let us divide our light
board in three parts
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to solve the three
little exercises.
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First, consider 2 squared
times 2 to the minus 1 over 2.
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Then you see you
have the same base.
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And therefore, to
get the product,
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it is sufficient just to
sum the two exponents.
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2 plus minus 1 over 2, which
is equal to 2 with exponent 3
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over 2.
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Now, I can rewrite this one
as 2 to 1 plus 1 over 2,
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which is equal 2 to 1 2
with the exponent 1 over 2.
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And this is, of course, equal
to 2 times square root of 2.
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And now, consider the
second expression.
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Then we have 3 cubed minus 5
squared over 2 to the 5 times 2
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to minus cube.
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OK, now, regarding the
numerator we just computed,
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these two numbers.
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And we get 27 minus 25 over--
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here, again, you
have the same base
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and then you can just
add the two exponents--
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2 to 5 minus 3, which is
equal to 2 over 2 squared.
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Again, this can be written
like 2 times 2 to minus 2,
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which is equal, because here we
have a product between powers
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with the same base.
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And therefore, we can just
add the two exponents.
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We have 2 minus 1, which
is equal to 1 over 2.
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And finally, consider
the last expression.
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2 cubed with the
exponent 2 cubed over 2 3
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squared with exponent 3.
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Now, when you have
the power of a power,
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you have to multiply
the exponents.
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Therefore, this is equal to
2 cube times 2 cubed over 2 3
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squared times 3.
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And this is exactly like to
write 2 3 times 2 cubed times 2
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to minus 3 cubed times 3.
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Now, again, we
have the same base.
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And therefore, we can just
add the two exponents.
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OK, maybe we can just
collect one copy of 3.
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And then we have 2 with
the exponent 3 times
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2 cubed minus 3 squared,
which is equal 2 cubed times--
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OK, 2 cubed is equal to 8.
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3 squared is equal to 9.
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Therefore, 8 minus 9, which
is equal to 2 minus 3,
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which is 1 over 2 to the
3rd, which is 1 over 8.
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Thank you for your attention.
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Hello.
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Let us consider
the exercise number
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2 of the rational
powers in practice step.
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OK, we have two real
numbers, y and x,
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with y greater or
equal than x, which
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is greater or equal than 0.
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And we want to compare these
two quantities, 2 times x to 3
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over 5, and this expression.
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OK, let us try to
solve this exercise.
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We have to ask ourselves, if
2 times x with the exponent 3
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over 5, is less or
equal or greater
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than y times y with
exponent minus 2 over 5
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plus y with exponent of minus 3
over 5 times the 5th root of y.
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OK, this second
expression is equal to y
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times y minus 2 over 5
plus y y minus 3 over 5
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times the 5th root of y.
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How can we rewrite this?
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OK, y to 1 is exactly like
to write y to 5 over 5.
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And I repeat, y to minus 2
over 5 plus, again, y to 5
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over 5 y minus 3 over 5.
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And now, the 5th root of y
is exactly y to 1 over 5.
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Here, we have a product of
two powers with the same base.
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Therefore, it is sufficient
just to add the two exponents,
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and we get y to 3 over 5.
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Again here, we have the y to
2 over 5 times y 1 over 5.
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But again, here we
have the same base.
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And then we get y to 3 over 5,
which is 2 times y to 3 over 5.
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OK, now, you see,
the function which
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sends z to z with the
exponent of 3 over 5
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is defined on
positive real numbers,
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and is strictly increasing.
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Therefore, x to 3 over 5 is
surely less or equal than y
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to 3 over 5 under
our hypothesis.
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But also, 2 times x to 3 over 5
is less or equal to 2 times y 3
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over 5.
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Therefore, which is the
disequality between the two
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initial quantities.
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Observe that, of
course, for x equal y,
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these two quantities are equal.
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And otherwise, the
quantity on the left
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is less than the
quantity on the right.
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Therefore, we can conclude
that the relation of it,
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between the two
initial quantities,
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is the following one.
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2 times x to 3 over 5 is less
or equal than this expression.
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