-0.000 000 000 742 147 677 42 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 42(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 42(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 42| = 0.000 000 000 742 147 677 42


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 42.

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 42 × 2 = 0 + 0.000 000 001 484 295 354 84;
  • 2) 0.000 000 001 484 295 354 84 × 2 = 0 + 0.000 000 002 968 590 709 68;
  • 3) 0.000 000 002 968 590 709 68 × 2 = 0 + 0.000 000 005 937 181 419 36;
  • 4) 0.000 000 005 937 181 419 36 × 2 = 0 + 0.000 000 011 874 362 838 72;
  • 5) 0.000 000 011 874 362 838 72 × 2 = 0 + 0.000 000 023 748 725 677 44;
  • 6) 0.000 000 023 748 725 677 44 × 2 = 0 + 0.000 000 047 497 451 354 88;
  • 7) 0.000 000 047 497 451 354 88 × 2 = 0 + 0.000 000 094 994 902 709 76;
  • 8) 0.000 000 094 994 902 709 76 × 2 = 0 + 0.000 000 189 989 805 419 52;
  • 9) 0.000 000 189 989 805 419 52 × 2 = 0 + 0.000 000 379 979 610 839 04;
  • 10) 0.000 000 379 979 610 839 04 × 2 = 0 + 0.000 000 759 959 221 678 08;
  • 11) 0.000 000 759 959 221 678 08 × 2 = 0 + 0.000 001 519 918 443 356 16;
  • 12) 0.000 001 519 918 443 356 16 × 2 = 0 + 0.000 003 039 836 886 712 32;
  • 13) 0.000 003 039 836 886 712 32 × 2 = 0 + 0.000 006 079 673 773 424 64;
  • 14) 0.000 006 079 673 773 424 64 × 2 = 0 + 0.000 012 159 347 546 849 28;
  • 15) 0.000 012 159 347 546 849 28 × 2 = 0 + 0.000 024 318 695 093 698 56;
  • 16) 0.000 024 318 695 093 698 56 × 2 = 0 + 0.000 048 637 390 187 397 12;
  • 17) 0.000 048 637 390 187 397 12 × 2 = 0 + 0.000 097 274 780 374 794 24;
  • 18) 0.000 097 274 780 374 794 24 × 2 = 0 + 0.000 194 549 560 749 588 48;
  • 19) 0.000 194 549 560 749 588 48 × 2 = 0 + 0.000 389 099 121 499 176 96;
  • 20) 0.000 389 099 121 499 176 96 × 2 = 0 + 0.000 778 198 242 998 353 92;
  • 21) 0.000 778 198 242 998 353 92 × 2 = 0 + 0.001 556 396 485 996 707 84;
  • 22) 0.001 556 396 485 996 707 84 × 2 = 0 + 0.003 112 792 971 993 415 68;
  • 23) 0.003 112 792 971 993 415 68 × 2 = 0 + 0.006 225 585 943 986 831 36;
  • 24) 0.006 225 585 943 986 831 36 × 2 = 0 + 0.012 451 171 887 973 662 72;
  • 25) 0.012 451 171 887 973 662 72 × 2 = 0 + 0.024 902 343 775 947 325 44;
  • 26) 0.024 902 343 775 947 325 44 × 2 = 0 + 0.049 804 687 551 894 650 88;
  • 27) 0.049 804 687 551 894 650 88 × 2 = 0 + 0.099 609 375 103 789 301 76;
  • 28) 0.099 609 375 103 789 301 76 × 2 = 0 + 0.199 218 750 207 578 603 52;
  • 29) 0.199 218 750 207 578 603 52 × 2 = 0 + 0.398 437 500 415 157 207 04;
  • 30) 0.398 437 500 415 157 207 04 × 2 = 0 + 0.796 875 000 830 314 414 08;
  • 31) 0.796 875 000 830 314 414 08 × 2 = 1 + 0.593 750 001 660 628 828 16;
  • 32) 0.593 750 001 660 628 828 16 × 2 = 1 + 0.187 500 003 321 257 656 32;
  • 33) 0.187 500 003 321 257 656 32 × 2 = 0 + 0.375 000 006 642 515 312 64;
  • 34) 0.375 000 006 642 515 312 64 × 2 = 0 + 0.750 000 013 285 030 625 28;
  • 35) 0.750 000 013 285 030 625 28 × 2 = 1 + 0.500 000 026 570 061 250 56;
  • 36) 0.500 000 026 570 061 250 56 × 2 = 1 + 0.000 000 053 140 122 501 12;
  • 37) 0.000 000 053 140 122 501 12 × 2 = 0 + 0.000 000 106 280 245 002 24;
  • 38) 0.000 000 106 280 245 002 24 × 2 = 0 + 0.000 000 212 560 490 004 48;
  • 39) 0.000 000 212 560 490 004 48 × 2 = 0 + 0.000 000 425 120 980 008 96;
  • 40) 0.000 000 425 120 980 008 96 × 2 = 0 + 0.000 000 850 241 960 017 92;
  • 41) 0.000 000 850 241 960 017 92 × 2 = 0 + 0.000 001 700 483 920 035 84;
  • 42) 0.000 001 700 483 920 035 84 × 2 = 0 + 0.000 003 400 967 840 071 68;
  • 43) 0.000 003 400 967 840 071 68 × 2 = 0 + 0.000 006 801 935 680 143 36;
  • 44) 0.000 006 801 935 680 143 36 × 2 = 0 + 0.000 013 603 871 360 286 72;
  • 45) 0.000 013 603 871 360 286 72 × 2 = 0 + 0.000 027 207 742 720 573 44;
  • 46) 0.000 027 207 742 720 573 44 × 2 = 0 + 0.000 054 415 485 441 146 88;
  • 47) 0.000 054 415 485 441 146 88 × 2 = 0 + 0.000 108 830 970 882 293 76;
  • 48) 0.000 108 830 970 882 293 76 × 2 = 0 + 0.000 217 661 941 764 587 52;
  • 49) 0.000 217 661 941 764 587 52 × 2 = 0 + 0.000 435 323 883 529 175 04;
  • 50) 0.000 435 323 883 529 175 04 × 2 = 0 + 0.000 870 647 767 058 350 08;
  • 51) 0.000 870 647 767 058 350 08 × 2 = 0 + 0.001 741 295 534 116 700 16;
  • 52) 0.001 741 295 534 116 700 16 × 2 = 0 + 0.003 482 591 068 233 400 32;
  • 53) 0.003 482 591 068 233 400 32 × 2 = 0 + 0.006 965 182 136 466 800 64;
  • 54) 0.006 965 182 136 466 800 64 × 2 = 0 + 0.013 930 364 272 933 601 28;

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 42(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 42(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 42(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 42 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