-0.000 000 000 742 147 677 57 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 57(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
-0.000 000 000 742 147 677 57(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:

|-0.000 000 000 742 147 677 57| = 0.000 000 000 742 147 677 57


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: 0.000 000 000 742 147 677 57.

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) 0.000 000 000 742 147 677 57 × 2 = 0 + 0.000 000 001 484 295 355 14;
  • 2) 0.000 000 001 484 295 355 14 × 2 = 0 + 0.000 000 002 968 590 710 28;
  • 3) 0.000 000 002 968 590 710 28 × 2 = 0 + 0.000 000 005 937 181 420 56;
  • 4) 0.000 000 005 937 181 420 56 × 2 = 0 + 0.000 000 011 874 362 841 12;
  • 5) 0.000 000 011 874 362 841 12 × 2 = 0 + 0.000 000 023 748 725 682 24;
  • 6) 0.000 000 023 748 725 682 24 × 2 = 0 + 0.000 000 047 497 451 364 48;
  • 7) 0.000 000 047 497 451 364 48 × 2 = 0 + 0.000 000 094 994 902 728 96;
  • 8) 0.000 000 094 994 902 728 96 × 2 = 0 + 0.000 000 189 989 805 457 92;
  • 9) 0.000 000 189 989 805 457 92 × 2 = 0 + 0.000 000 379 979 610 915 84;
  • 10) 0.000 000 379 979 610 915 84 × 2 = 0 + 0.000 000 759 959 221 831 68;
  • 11) 0.000 000 759 959 221 831 68 × 2 = 0 + 0.000 001 519 918 443 663 36;
  • 12) 0.000 001 519 918 443 663 36 × 2 = 0 + 0.000 003 039 836 887 326 72;
  • 13) 0.000 003 039 836 887 326 72 × 2 = 0 + 0.000 006 079 673 774 653 44;
  • 14) 0.000 006 079 673 774 653 44 × 2 = 0 + 0.000 012 159 347 549 306 88;
  • 15) 0.000 012 159 347 549 306 88 × 2 = 0 + 0.000 024 318 695 098 613 76;
  • 16) 0.000 024 318 695 098 613 76 × 2 = 0 + 0.000 048 637 390 197 227 52;
  • 17) 0.000 048 637 390 197 227 52 × 2 = 0 + 0.000 097 274 780 394 455 04;
  • 18) 0.000 097 274 780 394 455 04 × 2 = 0 + 0.000 194 549 560 788 910 08;
  • 19) 0.000 194 549 560 788 910 08 × 2 = 0 + 0.000 389 099 121 577 820 16;
  • 20) 0.000 389 099 121 577 820 16 × 2 = 0 + 0.000 778 198 243 155 640 32;
  • 21) 0.000 778 198 243 155 640 32 × 2 = 0 + 0.001 556 396 486 311 280 64;
  • 22) 0.001 556 396 486 311 280 64 × 2 = 0 + 0.003 112 792 972 622 561 28;
  • 23) 0.003 112 792 972 622 561 28 × 2 = 0 + 0.006 225 585 945 245 122 56;
  • 24) 0.006 225 585 945 245 122 56 × 2 = 0 + 0.012 451 171 890 490 245 12;
  • 25) 0.012 451 171 890 490 245 12 × 2 = 0 + 0.024 902 343 780 980 490 24;
  • 26) 0.024 902 343 780 980 490 24 × 2 = 0 + 0.049 804 687 561 960 980 48;
  • 27) 0.049 804 687 561 960 980 48 × 2 = 0 + 0.099 609 375 123 921 960 96;
  • 28) 0.099 609 375 123 921 960 96 × 2 = 0 + 0.199 218 750 247 843 921 92;
  • 29) 0.199 218 750 247 843 921 92 × 2 = 0 + 0.398 437 500 495 687 843 84;
  • 30) 0.398 437 500 495 687 843 84 × 2 = 0 + 0.796 875 000 991 375 687 68;
  • 31) 0.796 875 000 991 375 687 68 × 2 = 1 + 0.593 750 001 982 751 375 36;
  • 32) 0.593 750 001 982 751 375 36 × 2 = 1 + 0.187 500 003 965 502 750 72;
  • 33) 0.187 500 003 965 502 750 72 × 2 = 0 + 0.375 000 007 931 005 501 44;
  • 34) 0.375 000 007 931 005 501 44 × 2 = 0 + 0.750 000 015 862 011 002 88;
  • 35) 0.750 000 015 862 011 002 88 × 2 = 1 + 0.500 000 031 724 022 005 76;
  • 36) 0.500 000 031 724 022 005 76 × 2 = 1 + 0.000 000 063 448 044 011 52;
  • 37) 0.000 000 063 448 044 011 52 × 2 = 0 + 0.000 000 126 896 088 023 04;
  • 38) 0.000 000 126 896 088 023 04 × 2 = 0 + 0.000 000 253 792 176 046 08;
  • 39) 0.000 000 253 792 176 046 08 × 2 = 0 + 0.000 000 507 584 352 092 16;
  • 40) 0.000 000 507 584 352 092 16 × 2 = 0 + 0.000 001 015 168 704 184 32;
  • 41) 0.000 001 015 168 704 184 32 × 2 = 0 + 0.000 002 030 337 408 368 64;
  • 42) 0.000 002 030 337 408 368 64 × 2 = 0 + 0.000 004 060 674 816 737 28;
  • 43) 0.000 004 060 674 816 737 28 × 2 = 0 + 0.000 008 121 349 633 474 56;
  • 44) 0.000 008 121 349 633 474 56 × 2 = 0 + 0.000 016 242 699 266 949 12;
  • 45) 0.000 016 242 699 266 949 12 × 2 = 0 + 0.000 032 485 398 533 898 24;
  • 46) 0.000 032 485 398 533 898 24 × 2 = 0 + 0.000 064 970 797 067 796 48;
  • 47) 0.000 064 970 797 067 796 48 × 2 = 0 + 0.000 129 941 594 135 592 96;
  • 48) 0.000 129 941 594 135 592 96 × 2 = 0 + 0.000 259 883 188 271 185 92;
  • 49) 0.000 259 883 188 271 185 92 × 2 = 0 + 0.000 519 766 376 542 371 84;
  • 50) 0.000 519 766 376 542 371 84 × 2 = 0 + 0.001 039 532 753 084 743 68;
  • 51) 0.001 039 532 753 084 743 68 × 2 = 0 + 0.002 079 065 506 169 487 36;
  • 52) 0.002 079 065 506 169 487 36 × 2 = 0 + 0.004 158 131 012 338 974 72;
  • 53) 0.004 158 131 012 338 974 72 × 2 = 0 + 0.008 316 262 024 677 949 44;
  • 54) 0.008 316 262 024 677 949 44 × 2 = 0 + 0.016 632 524 049 355 898 88;

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:


0.000 000 000 742 147 677 57(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 677 57(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:


0.000 000 000 742 147 677 57(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 -0.000 000 000 742 147 677 57 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