-0.000 000 000 740 16 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 740 16(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
-0.000 000 000 740 16(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:

|-0.000 000 000 740 16| = 0.000 000 000 740 16


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: 0.000 000 000 740 16.

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) 0.000 000 000 740 16 × 2 = 0 + 0.000 000 001 480 32;
  • 2) 0.000 000 001 480 32 × 2 = 0 + 0.000 000 002 960 64;
  • 3) 0.000 000 002 960 64 × 2 = 0 + 0.000 000 005 921 28;
  • 4) 0.000 000 005 921 28 × 2 = 0 + 0.000 000 011 842 56;
  • 5) 0.000 000 011 842 56 × 2 = 0 + 0.000 000 023 685 12;
  • 6) 0.000 000 023 685 12 × 2 = 0 + 0.000 000 047 370 24;
  • 7) 0.000 000 047 370 24 × 2 = 0 + 0.000 000 094 740 48;
  • 8) 0.000 000 094 740 48 × 2 = 0 + 0.000 000 189 480 96;
  • 9) 0.000 000 189 480 96 × 2 = 0 + 0.000 000 378 961 92;
  • 10) 0.000 000 378 961 92 × 2 = 0 + 0.000 000 757 923 84;
  • 11) 0.000 000 757 923 84 × 2 = 0 + 0.000 001 515 847 68;
  • 12) 0.000 001 515 847 68 × 2 = 0 + 0.000 003 031 695 36;
  • 13) 0.000 003 031 695 36 × 2 = 0 + 0.000 006 063 390 72;
  • 14) 0.000 006 063 390 72 × 2 = 0 + 0.000 012 126 781 44;
  • 15) 0.000 012 126 781 44 × 2 = 0 + 0.000 024 253 562 88;
  • 16) 0.000 024 253 562 88 × 2 = 0 + 0.000 048 507 125 76;
  • 17) 0.000 048 507 125 76 × 2 = 0 + 0.000 097 014 251 52;
  • 18) 0.000 097 014 251 52 × 2 = 0 + 0.000 194 028 503 04;
  • 19) 0.000 194 028 503 04 × 2 = 0 + 0.000 388 057 006 08;
  • 20) 0.000 388 057 006 08 × 2 = 0 + 0.000 776 114 012 16;
  • 21) 0.000 776 114 012 16 × 2 = 0 + 0.001 552 228 024 32;
  • 22) 0.001 552 228 024 32 × 2 = 0 + 0.003 104 456 048 64;
  • 23) 0.003 104 456 048 64 × 2 = 0 + 0.006 208 912 097 28;
  • 24) 0.006 208 912 097 28 × 2 = 0 + 0.012 417 824 194 56;
  • 25) 0.012 417 824 194 56 × 2 = 0 + 0.024 835 648 389 12;
  • 26) 0.024 835 648 389 12 × 2 = 0 + 0.049 671 296 778 24;
  • 27) 0.049 671 296 778 24 × 2 = 0 + 0.099 342 593 556 48;
  • 28) 0.099 342 593 556 48 × 2 = 0 + 0.198 685 187 112 96;
  • 29) 0.198 685 187 112 96 × 2 = 0 + 0.397 370 374 225 92;
  • 30) 0.397 370 374 225 92 × 2 = 0 + 0.794 740 748 451 84;
  • 31) 0.794 740 748 451 84 × 2 = 1 + 0.589 481 496 903 68;
  • 32) 0.589 481 496 903 68 × 2 = 1 + 0.178 962 993 807 36;
  • 33) 0.178 962 993 807 36 × 2 = 0 + 0.357 925 987 614 72;
  • 34) 0.357 925 987 614 72 × 2 = 0 + 0.715 851 975 229 44;
  • 35) 0.715 851 975 229 44 × 2 = 1 + 0.431 703 950 458 88;
  • 36) 0.431 703 950 458 88 × 2 = 0 + 0.863 407 900 917 76;
  • 37) 0.863 407 900 917 76 × 2 = 1 + 0.726 815 801 835 52;
  • 38) 0.726 815 801 835 52 × 2 = 1 + 0.453 631 603 671 04;
  • 39) 0.453 631 603 671 04 × 2 = 0 + 0.907 263 207 342 08;
  • 40) 0.907 263 207 342 08 × 2 = 1 + 0.814 526 414 684 16;
  • 41) 0.814 526 414 684 16 × 2 = 1 + 0.629 052 829 368 32;
  • 42) 0.629 052 829 368 32 × 2 = 1 + 0.258 105 658 736 64;
  • 43) 0.258 105 658 736 64 × 2 = 0 + 0.516 211 317 473 28;
  • 44) 0.516 211 317 473 28 × 2 = 1 + 0.032 422 634 946 56;
  • 45) 0.032 422 634 946 56 × 2 = 0 + 0.064 845 269 893 12;
  • 46) 0.064 845 269 893 12 × 2 = 0 + 0.129 690 539 786 24;
  • 47) 0.129 690 539 786 24 × 2 = 0 + 0.259 381 079 572 48;
  • 48) 0.259 381 079 572 48 × 2 = 0 + 0.518 762 159 144 96;
  • 49) 0.518 762 159 144 96 × 2 = 1 + 0.037 524 318 289 92;
  • 50) 0.037 524 318 289 92 × 2 = 0 + 0.075 048 636 579 84;
  • 51) 0.075 048 636 579 84 × 2 = 0 + 0.150 097 273 159 68;
  • 52) 0.150 097 273 159 68 × 2 = 0 + 0.300 194 546 319 36;
  • 53) 0.300 194 546 319 36 × 2 = 0 + 0.600 389 092 638 72;
  • 54) 0.600 389 092 638 72 × 2 = 1 + 0.200 778 185 277 44;

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:


0.000 000 000 740 16(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1101 0000 1000 01(2)

6. Positive number before normalization:

0.000 000 000 740 16(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1101 0000 1000 01(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:


0.000 000 000 740 16(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1101 0000 1000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1101 0000 1000 01(2) × 20 =

1.1001 0110 1110 1000 0100 001(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 0110 1110 1000 0100 001


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 0111 0100 0010 0001 =

100 1011 0111 0100 0010 0001


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 0111 0100 0010 0001


Decimal number -0.000 000 000 740 16 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 0111 0100 0010 0001

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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