-0.000 000 000 741 99 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 99(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 741 99(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 741 99| = 0.000 000 000 741 99


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 741 99.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 741 99 × 2 = 0 + 0.000 000 001 483 98;
  • 2) 0.000 000 001 483 98 × 2 = 0 + 0.000 000 002 967 96;
  • 3) 0.000 000 002 967 96 × 2 = 0 + 0.000 000 005 935 92;
  • 4) 0.000 000 005 935 92 × 2 = 0 + 0.000 000 011 871 84;
  • 5) 0.000 000 011 871 84 × 2 = 0 + 0.000 000 023 743 68;
  • 6) 0.000 000 023 743 68 × 2 = 0 + 0.000 000 047 487 36;
  • 7) 0.000 000 047 487 36 × 2 = 0 + 0.000 000 094 974 72;
  • 8) 0.000 000 094 974 72 × 2 = 0 + 0.000 000 189 949 44;
  • 9) 0.000 000 189 949 44 × 2 = 0 + 0.000 000 379 898 88;
  • 10) 0.000 000 379 898 88 × 2 = 0 + 0.000 000 759 797 76;
  • 11) 0.000 000 759 797 76 × 2 = 0 + 0.000 001 519 595 52;
  • 12) 0.000 001 519 595 52 × 2 = 0 + 0.000 003 039 191 04;
  • 13) 0.000 003 039 191 04 × 2 = 0 + 0.000 006 078 382 08;
  • 14) 0.000 006 078 382 08 × 2 = 0 + 0.000 012 156 764 16;
  • 15) 0.000 012 156 764 16 × 2 = 0 + 0.000 024 313 528 32;
  • 16) 0.000 024 313 528 32 × 2 = 0 + 0.000 048 627 056 64;
  • 17) 0.000 048 627 056 64 × 2 = 0 + 0.000 097 254 113 28;
  • 18) 0.000 097 254 113 28 × 2 = 0 + 0.000 194 508 226 56;
  • 19) 0.000 194 508 226 56 × 2 = 0 + 0.000 389 016 453 12;
  • 20) 0.000 389 016 453 12 × 2 = 0 + 0.000 778 032 906 24;
  • 21) 0.000 778 032 906 24 × 2 = 0 + 0.001 556 065 812 48;
  • 22) 0.001 556 065 812 48 × 2 = 0 + 0.003 112 131 624 96;
  • 23) 0.003 112 131 624 96 × 2 = 0 + 0.006 224 263 249 92;
  • 24) 0.006 224 263 249 92 × 2 = 0 + 0.012 448 526 499 84;
  • 25) 0.012 448 526 499 84 × 2 = 0 + 0.024 897 052 999 68;
  • 26) 0.024 897 052 999 68 × 2 = 0 + 0.049 794 105 999 36;
  • 27) 0.049 794 105 999 36 × 2 = 0 + 0.099 588 211 998 72;
  • 28) 0.099 588 211 998 72 × 2 = 0 + 0.199 176 423 997 44;
  • 29) 0.199 176 423 997 44 × 2 = 0 + 0.398 352 847 994 88;
  • 30) 0.398 352 847 994 88 × 2 = 0 + 0.796 705 695 989 76;
  • 31) 0.796 705 695 989 76 × 2 = 1 + 0.593 411 391 979 52;
  • 32) 0.593 411 391 979 52 × 2 = 1 + 0.186 822 783 959 04;
  • 33) 0.186 822 783 959 04 × 2 = 0 + 0.373 645 567 918 08;
  • 34) 0.373 645 567 918 08 × 2 = 0 + 0.747 291 135 836 16;
  • 35) 0.747 291 135 836 16 × 2 = 1 + 0.494 582 271 672 32;
  • 36) 0.494 582 271 672 32 × 2 = 0 + 0.989 164 543 344 64;
  • 37) 0.989 164 543 344 64 × 2 = 1 + 0.978 329 086 689 28;
  • 38) 0.978 329 086 689 28 × 2 = 1 + 0.956 658 173 378 56;
  • 39) 0.956 658 173 378 56 × 2 = 1 + 0.913 316 346 757 12;
  • 40) 0.913 316 346 757 12 × 2 = 1 + 0.826 632 693 514 24;
  • 41) 0.826 632 693 514 24 × 2 = 1 + 0.653 265 387 028 48;
  • 42) 0.653 265 387 028 48 × 2 = 1 + 0.306 530 774 056 96;
  • 43) 0.306 530 774 056 96 × 2 = 0 + 0.613 061 548 113 92;
  • 44) 0.613 061 548 113 92 × 2 = 1 + 0.226 123 096 227 84;
  • 45) 0.226 123 096 227 84 × 2 = 0 + 0.452 246 192 455 68;
  • 46) 0.452 246 192 455 68 × 2 = 0 + 0.904 492 384 911 36;
  • 47) 0.904 492 384 911 36 × 2 = 1 + 0.808 984 769 822 72;
  • 48) 0.808 984 769 822 72 × 2 = 1 + 0.617 969 539 645 44;
  • 49) 0.617 969 539 645 44 × 2 = 1 + 0.235 939 079 290 88;
  • 50) 0.235 939 079 290 88 × 2 = 0 + 0.471 878 158 581 76;
  • 51) 0.471 878 158 581 76 × 2 = 0 + 0.943 756 317 163 52;
  • 52) 0.943 756 317 163 52 × 2 = 1 + 0.887 512 634 327 04;
  • 53) 0.887 512 634 327 04 × 2 = 1 + 0.775 025 268 654 08;
  • 54) 0.775 025 268 654 08 × 2 = 1 + 0.550 050 537 308 16;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 741 99(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0011 1001 11(2)

6. Positive number before normalization:

0.000 000 000 741 99(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0011 1001 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 741 99(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0011 1001 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0011 1001 11(2) × 20 =

1.1001 0111 1110 1001 1100 111(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1110 1001 1100 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 0100 1110 0111 =

100 1011 1111 0100 1110 0111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 0100 1110 0111


Decimal number -0.000 000 000 741 99 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 0100 1110 0111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111