-0.000 000 000 742 147 678 03 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 678 03(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
-0.000 000 000 742 147 678 03(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:

|-0.000 000 000 742 147 678 03| = 0.000 000 000 742 147 678 03


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: 0.000 000 000 742 147 678 03.

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) 0.000 000 000 742 147 678 03 × 2 = 0 + 0.000 000 001 484 295 356 06;
  • 2) 0.000 000 001 484 295 356 06 × 2 = 0 + 0.000 000 002 968 590 712 12;
  • 3) 0.000 000 002 968 590 712 12 × 2 = 0 + 0.000 000 005 937 181 424 24;
  • 4) 0.000 000 005 937 181 424 24 × 2 = 0 + 0.000 000 011 874 362 848 48;
  • 5) 0.000 000 011 874 362 848 48 × 2 = 0 + 0.000 000 023 748 725 696 96;
  • 6) 0.000 000 023 748 725 696 96 × 2 = 0 + 0.000 000 047 497 451 393 92;
  • 7) 0.000 000 047 497 451 393 92 × 2 = 0 + 0.000 000 094 994 902 787 84;
  • 8) 0.000 000 094 994 902 787 84 × 2 = 0 + 0.000 000 189 989 805 575 68;
  • 9) 0.000 000 189 989 805 575 68 × 2 = 0 + 0.000 000 379 979 611 151 36;
  • 10) 0.000 000 379 979 611 151 36 × 2 = 0 + 0.000 000 759 959 222 302 72;
  • 11) 0.000 000 759 959 222 302 72 × 2 = 0 + 0.000 001 519 918 444 605 44;
  • 12) 0.000 001 519 918 444 605 44 × 2 = 0 + 0.000 003 039 836 889 210 88;
  • 13) 0.000 003 039 836 889 210 88 × 2 = 0 + 0.000 006 079 673 778 421 76;
  • 14) 0.000 006 079 673 778 421 76 × 2 = 0 + 0.000 012 159 347 556 843 52;
  • 15) 0.000 012 159 347 556 843 52 × 2 = 0 + 0.000 024 318 695 113 687 04;
  • 16) 0.000 024 318 695 113 687 04 × 2 = 0 + 0.000 048 637 390 227 374 08;
  • 17) 0.000 048 637 390 227 374 08 × 2 = 0 + 0.000 097 274 780 454 748 16;
  • 18) 0.000 097 274 780 454 748 16 × 2 = 0 + 0.000 194 549 560 909 496 32;
  • 19) 0.000 194 549 560 909 496 32 × 2 = 0 + 0.000 389 099 121 818 992 64;
  • 20) 0.000 389 099 121 818 992 64 × 2 = 0 + 0.000 778 198 243 637 985 28;
  • 21) 0.000 778 198 243 637 985 28 × 2 = 0 + 0.001 556 396 487 275 970 56;
  • 22) 0.001 556 396 487 275 970 56 × 2 = 0 + 0.003 112 792 974 551 941 12;
  • 23) 0.003 112 792 974 551 941 12 × 2 = 0 + 0.006 225 585 949 103 882 24;
  • 24) 0.006 225 585 949 103 882 24 × 2 = 0 + 0.012 451 171 898 207 764 48;
  • 25) 0.012 451 171 898 207 764 48 × 2 = 0 + 0.024 902 343 796 415 528 96;
  • 26) 0.024 902 343 796 415 528 96 × 2 = 0 + 0.049 804 687 592 831 057 92;
  • 27) 0.049 804 687 592 831 057 92 × 2 = 0 + 0.099 609 375 185 662 115 84;
  • 28) 0.099 609 375 185 662 115 84 × 2 = 0 + 0.199 218 750 371 324 231 68;
  • 29) 0.199 218 750 371 324 231 68 × 2 = 0 + 0.398 437 500 742 648 463 36;
  • 30) 0.398 437 500 742 648 463 36 × 2 = 0 + 0.796 875 001 485 296 926 72;
  • 31) 0.796 875 001 485 296 926 72 × 2 = 1 + 0.593 750 002 970 593 853 44;
  • 32) 0.593 750 002 970 593 853 44 × 2 = 1 + 0.187 500 005 941 187 706 88;
  • 33) 0.187 500 005 941 187 706 88 × 2 = 0 + 0.375 000 011 882 375 413 76;
  • 34) 0.375 000 011 882 375 413 76 × 2 = 0 + 0.750 000 023 764 750 827 52;
  • 35) 0.750 000 023 764 750 827 52 × 2 = 1 + 0.500 000 047 529 501 655 04;
  • 36) 0.500 000 047 529 501 655 04 × 2 = 1 + 0.000 000 095 059 003 310 08;
  • 37) 0.000 000 095 059 003 310 08 × 2 = 0 + 0.000 000 190 118 006 620 16;
  • 38) 0.000 000 190 118 006 620 16 × 2 = 0 + 0.000 000 380 236 013 240 32;
  • 39) 0.000 000 380 236 013 240 32 × 2 = 0 + 0.000 000 760 472 026 480 64;
  • 40) 0.000 000 760 472 026 480 64 × 2 = 0 + 0.000 001 520 944 052 961 28;
  • 41) 0.000 001 520 944 052 961 28 × 2 = 0 + 0.000 003 041 888 105 922 56;
  • 42) 0.000 003 041 888 105 922 56 × 2 = 0 + 0.000 006 083 776 211 845 12;
  • 43) 0.000 006 083 776 211 845 12 × 2 = 0 + 0.000 012 167 552 423 690 24;
  • 44) 0.000 012 167 552 423 690 24 × 2 = 0 + 0.000 024 335 104 847 380 48;
  • 45) 0.000 024 335 104 847 380 48 × 2 = 0 + 0.000 048 670 209 694 760 96;
  • 46) 0.000 048 670 209 694 760 96 × 2 = 0 + 0.000 097 340 419 389 521 92;
  • 47) 0.000 097 340 419 389 521 92 × 2 = 0 + 0.000 194 680 838 779 043 84;
  • 48) 0.000 194 680 838 779 043 84 × 2 = 0 + 0.000 389 361 677 558 087 68;
  • 49) 0.000 389 361 677 558 087 68 × 2 = 0 + 0.000 778 723 355 116 175 36;
  • 50) 0.000 778 723 355 116 175 36 × 2 = 0 + 0.001 557 446 710 232 350 72;
  • 51) 0.001 557 446 710 232 350 72 × 2 = 0 + 0.003 114 893 420 464 701 44;
  • 52) 0.003 114 893 420 464 701 44 × 2 = 0 + 0.006 229 786 840 929 402 88;
  • 53) 0.006 229 786 840 929 402 88 × 2 = 0 + 0.012 459 573 681 858 805 76;
  • 54) 0.012 459 573 681 858 805 76 × 2 = 0 + 0.024 919 147 363 717 611 52;

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:


0.000 000 000 742 147 678 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 678 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 678 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 678 03 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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