-0.000 000 000 742 147 676 309 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 309(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
-0.000 000 000 742 147 676 309(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:

|-0.000 000 000 742 147 676 309| = 0.000 000 000 742 147 676 309


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: 0.000 000 000 742 147 676 309.

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) 0.000 000 000 742 147 676 309 × 2 = 0 + 0.000 000 001 484 295 352 618;
  • 2) 0.000 000 001 484 295 352 618 × 2 = 0 + 0.000 000 002 968 590 705 236;
  • 3) 0.000 000 002 968 590 705 236 × 2 = 0 + 0.000 000 005 937 181 410 472;
  • 4) 0.000 000 005 937 181 410 472 × 2 = 0 + 0.000 000 011 874 362 820 944;
  • 5) 0.000 000 011 874 362 820 944 × 2 = 0 + 0.000 000 023 748 725 641 888;
  • 6) 0.000 000 023 748 725 641 888 × 2 = 0 + 0.000 000 047 497 451 283 776;
  • 7) 0.000 000 047 497 451 283 776 × 2 = 0 + 0.000 000 094 994 902 567 552;
  • 8) 0.000 000 094 994 902 567 552 × 2 = 0 + 0.000 000 189 989 805 135 104;
  • 9) 0.000 000 189 989 805 135 104 × 2 = 0 + 0.000 000 379 979 610 270 208;
  • 10) 0.000 000 379 979 610 270 208 × 2 = 0 + 0.000 000 759 959 220 540 416;
  • 11) 0.000 000 759 959 220 540 416 × 2 = 0 + 0.000 001 519 918 441 080 832;
  • 12) 0.000 001 519 918 441 080 832 × 2 = 0 + 0.000 003 039 836 882 161 664;
  • 13) 0.000 003 039 836 882 161 664 × 2 = 0 + 0.000 006 079 673 764 323 328;
  • 14) 0.000 006 079 673 764 323 328 × 2 = 0 + 0.000 012 159 347 528 646 656;
  • 15) 0.000 012 159 347 528 646 656 × 2 = 0 + 0.000 024 318 695 057 293 312;
  • 16) 0.000 024 318 695 057 293 312 × 2 = 0 + 0.000 048 637 390 114 586 624;
  • 17) 0.000 048 637 390 114 586 624 × 2 = 0 + 0.000 097 274 780 229 173 248;
  • 18) 0.000 097 274 780 229 173 248 × 2 = 0 + 0.000 194 549 560 458 346 496;
  • 19) 0.000 194 549 560 458 346 496 × 2 = 0 + 0.000 389 099 120 916 692 992;
  • 20) 0.000 389 099 120 916 692 992 × 2 = 0 + 0.000 778 198 241 833 385 984;
  • 21) 0.000 778 198 241 833 385 984 × 2 = 0 + 0.001 556 396 483 666 771 968;
  • 22) 0.001 556 396 483 666 771 968 × 2 = 0 + 0.003 112 792 967 333 543 936;
  • 23) 0.003 112 792 967 333 543 936 × 2 = 0 + 0.006 225 585 934 667 087 872;
  • 24) 0.006 225 585 934 667 087 872 × 2 = 0 + 0.012 451 171 869 334 175 744;
  • 25) 0.012 451 171 869 334 175 744 × 2 = 0 + 0.024 902 343 738 668 351 488;
  • 26) 0.024 902 343 738 668 351 488 × 2 = 0 + 0.049 804 687 477 336 702 976;
  • 27) 0.049 804 687 477 336 702 976 × 2 = 0 + 0.099 609 374 954 673 405 952;
  • 28) 0.099 609 374 954 673 405 952 × 2 = 0 + 0.199 218 749 909 346 811 904;
  • 29) 0.199 218 749 909 346 811 904 × 2 = 0 + 0.398 437 499 818 693 623 808;
  • 30) 0.398 437 499 818 693 623 808 × 2 = 0 + 0.796 874 999 637 387 247 616;
  • 31) 0.796 874 999 637 387 247 616 × 2 = 1 + 0.593 749 999 274 774 495 232;
  • 32) 0.593 749 999 274 774 495 232 × 2 = 1 + 0.187 499 998 549 548 990 464;
  • 33) 0.187 499 998 549 548 990 464 × 2 = 0 + 0.374 999 997 099 097 980 928;
  • 34) 0.374 999 997 099 097 980 928 × 2 = 0 + 0.749 999 994 198 195 961 856;
  • 35) 0.749 999 994 198 195 961 856 × 2 = 1 + 0.499 999 988 396 391 923 712;
  • 36) 0.499 999 988 396 391 923 712 × 2 = 0 + 0.999 999 976 792 783 847 424;
  • 37) 0.999 999 976 792 783 847 424 × 2 = 1 + 0.999 999 953 585 567 694 848;
  • 38) 0.999 999 953 585 567 694 848 × 2 = 1 + 0.999 999 907 171 135 389 696;
  • 39) 0.999 999 907 171 135 389 696 × 2 = 1 + 0.999 999 814 342 270 779 392;
  • 40) 0.999 999 814 342 270 779 392 × 2 = 1 + 0.999 999 628 684 541 558 784;
  • 41) 0.999 999 628 684 541 558 784 × 2 = 1 + 0.999 999 257 369 083 117 568;
  • 42) 0.999 999 257 369 083 117 568 × 2 = 1 + 0.999 998 514 738 166 235 136;
  • 43) 0.999 998 514 738 166 235 136 × 2 = 1 + 0.999 997 029 476 332 470 272;
  • 44) 0.999 997 029 476 332 470 272 × 2 = 1 + 0.999 994 058 952 664 940 544;
  • 45) 0.999 994 058 952 664 940 544 × 2 = 1 + 0.999 988 117 905 329 881 088;
  • 46) 0.999 988 117 905 329 881 088 × 2 = 1 + 0.999 976 235 810 659 762 176;
  • 47) 0.999 976 235 810 659 762 176 × 2 = 1 + 0.999 952 471 621 319 524 352;
  • 48) 0.999 952 471 621 319 524 352 × 2 = 1 + 0.999 904 943 242 639 048 704;
  • 49) 0.999 904 943 242 639 048 704 × 2 = 1 + 0.999 809 886 485 278 097 408;
  • 50) 0.999 809 886 485 278 097 408 × 2 = 1 + 0.999 619 772 970 556 194 816;
  • 51) 0.999 619 772 970 556 194 816 × 2 = 1 + 0.999 239 545 941 112 389 632;
  • 52) 0.999 239 545 941 112 389 632 × 2 = 1 + 0.998 479 091 882 224 779 264;
  • 53) 0.998 479 091 882 224 779 264 × 2 = 1 + 0.996 958 183 764 449 558 528;
  • 54) 0.996 958 183 764 449 558 528 × 2 = 1 + 0.993 916 367 528 899 117 056;

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:


0.000 000 000 742 147 676 309(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 309(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 309(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 0111 1111 1111 1111 111


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 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 309 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 391 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