-0.000 000 000 742 147 676 646 796 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 796(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
-0.000 000 000 742 147 676 646 796(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:

|-0.000 000 000 742 147 676 646 796| = 0.000 000 000 742 147 676 646 796


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: 0.000 000 000 742 147 676 646 796.

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) 0.000 000 000 742 147 676 646 796 × 2 = 0 + 0.000 000 001 484 295 353 293 592;
  • 2) 0.000 000 001 484 295 353 293 592 × 2 = 0 + 0.000 000 002 968 590 706 587 184;
  • 3) 0.000 000 002 968 590 706 587 184 × 2 = 0 + 0.000 000 005 937 181 413 174 368;
  • 4) 0.000 000 005 937 181 413 174 368 × 2 = 0 + 0.000 000 011 874 362 826 348 736;
  • 5) 0.000 000 011 874 362 826 348 736 × 2 = 0 + 0.000 000 023 748 725 652 697 472;
  • 6) 0.000 000 023 748 725 652 697 472 × 2 = 0 + 0.000 000 047 497 451 305 394 944;
  • 7) 0.000 000 047 497 451 305 394 944 × 2 = 0 + 0.000 000 094 994 902 610 789 888;
  • 8) 0.000 000 094 994 902 610 789 888 × 2 = 0 + 0.000 000 189 989 805 221 579 776;
  • 9) 0.000 000 189 989 805 221 579 776 × 2 = 0 + 0.000 000 379 979 610 443 159 552;
  • 10) 0.000 000 379 979 610 443 159 552 × 2 = 0 + 0.000 000 759 959 220 886 319 104;
  • 11) 0.000 000 759 959 220 886 319 104 × 2 = 0 + 0.000 001 519 918 441 772 638 208;
  • 12) 0.000 001 519 918 441 772 638 208 × 2 = 0 + 0.000 003 039 836 883 545 276 416;
  • 13) 0.000 003 039 836 883 545 276 416 × 2 = 0 + 0.000 006 079 673 767 090 552 832;
  • 14) 0.000 006 079 673 767 090 552 832 × 2 = 0 + 0.000 012 159 347 534 181 105 664;
  • 15) 0.000 012 159 347 534 181 105 664 × 2 = 0 + 0.000 024 318 695 068 362 211 328;
  • 16) 0.000 024 318 695 068 362 211 328 × 2 = 0 + 0.000 048 637 390 136 724 422 656;
  • 17) 0.000 048 637 390 136 724 422 656 × 2 = 0 + 0.000 097 274 780 273 448 845 312;
  • 18) 0.000 097 274 780 273 448 845 312 × 2 = 0 + 0.000 194 549 560 546 897 690 624;
  • 19) 0.000 194 549 560 546 897 690 624 × 2 = 0 + 0.000 389 099 121 093 795 381 248;
  • 20) 0.000 389 099 121 093 795 381 248 × 2 = 0 + 0.000 778 198 242 187 590 762 496;
  • 21) 0.000 778 198 242 187 590 762 496 × 2 = 0 + 0.001 556 396 484 375 181 524 992;
  • 22) 0.001 556 396 484 375 181 524 992 × 2 = 0 + 0.003 112 792 968 750 363 049 984;
  • 23) 0.003 112 792 968 750 363 049 984 × 2 = 0 + 0.006 225 585 937 500 726 099 968;
  • 24) 0.006 225 585 937 500 726 099 968 × 2 = 0 + 0.012 451 171 875 001 452 199 936;
  • 25) 0.012 451 171 875 001 452 199 936 × 2 = 0 + 0.024 902 343 750 002 904 399 872;
  • 26) 0.024 902 343 750 002 904 399 872 × 2 = 0 + 0.049 804 687 500 005 808 799 744;
  • 27) 0.049 804 687 500 005 808 799 744 × 2 = 0 + 0.099 609 375 000 011 617 599 488;
  • 28) 0.099 609 375 000 011 617 599 488 × 2 = 0 + 0.199 218 750 000 023 235 198 976;
  • 29) 0.199 218 750 000 023 235 198 976 × 2 = 0 + 0.398 437 500 000 046 470 397 952;
  • 30) 0.398 437 500 000 046 470 397 952 × 2 = 0 + 0.796 875 000 000 092 940 795 904;
  • 31) 0.796 875 000 000 092 940 795 904 × 2 = 1 + 0.593 750 000 000 185 881 591 808;
  • 32) 0.593 750 000 000 185 881 591 808 × 2 = 1 + 0.187 500 000 000 371 763 183 616;
  • 33) 0.187 500 000 000 371 763 183 616 × 2 = 0 + 0.375 000 000 000 743 526 367 232;
  • 34) 0.375 000 000 000 743 526 367 232 × 2 = 0 + 0.750 000 000 001 487 052 734 464;
  • 35) 0.750 000 000 001 487 052 734 464 × 2 = 1 + 0.500 000 000 002 974 105 468 928;
  • 36) 0.500 000 000 002 974 105 468 928 × 2 = 1 + 0.000 000 000 005 948 210 937 856;
  • 37) 0.000 000 000 005 948 210 937 856 × 2 = 0 + 0.000 000 000 011 896 421 875 712;
  • 38) 0.000 000 000 011 896 421 875 712 × 2 = 0 + 0.000 000 000 023 792 843 751 424;
  • 39) 0.000 000 000 023 792 843 751 424 × 2 = 0 + 0.000 000 000 047 585 687 502 848;
  • 40) 0.000 000 000 047 585 687 502 848 × 2 = 0 + 0.000 000 000 095 171 375 005 696;
  • 41) 0.000 000 000 095 171 375 005 696 × 2 = 0 + 0.000 000 000 190 342 750 011 392;
  • 42) 0.000 000 000 190 342 750 011 392 × 2 = 0 + 0.000 000 000 380 685 500 022 784;
  • 43) 0.000 000 000 380 685 500 022 784 × 2 = 0 + 0.000 000 000 761 371 000 045 568;
  • 44) 0.000 000 000 761 371 000 045 568 × 2 = 0 + 0.000 000 001 522 742 000 091 136;
  • 45) 0.000 000 001 522 742 000 091 136 × 2 = 0 + 0.000 000 003 045 484 000 182 272;
  • 46) 0.000 000 003 045 484 000 182 272 × 2 = 0 + 0.000 000 006 090 968 000 364 544;
  • 47) 0.000 000 006 090 968 000 364 544 × 2 = 0 + 0.000 000 012 181 936 000 729 088;
  • 48) 0.000 000 012 181 936 000 729 088 × 2 = 0 + 0.000 000 024 363 872 001 458 176;
  • 49) 0.000 000 024 363 872 001 458 176 × 2 = 0 + 0.000 000 048 727 744 002 916 352;
  • 50) 0.000 000 048 727 744 002 916 352 × 2 = 0 + 0.000 000 097 455 488 005 832 704;
  • 51) 0.000 000 097 455 488 005 832 704 × 2 = 0 + 0.000 000 194 910 976 011 665 408;
  • 52) 0.000 000 194 910 976 011 665 408 × 2 = 0 + 0.000 000 389 821 952 023 330 816;
  • 53) 0.000 000 389 821 952 023 330 816 × 2 = 0 + 0.000 000 779 643 904 046 661 632;
  • 54) 0.000 000 779 643 904 046 661 632 × 2 = 0 + 0.000 001 559 287 808 093 323 264;

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:


0.000 000 000 742 147 676 646 796(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 796(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:


0.000 000 000 742 147 676 646 796(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 -0.000 000 000 742 147 676 646 796 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 646 857 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 minutes

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