-0.000 000 000 742 147 676 247 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 247(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
-0.000 000 000 742 147 676 247(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:

|-0.000 000 000 742 147 676 247| = 0.000 000 000 742 147 676 247


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: 0.000 000 000 742 147 676 247.

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) 0.000 000 000 742 147 676 247 × 2 = 0 + 0.000 000 001 484 295 352 494;
  • 2) 0.000 000 001 484 295 352 494 × 2 = 0 + 0.000 000 002 968 590 704 988;
  • 3) 0.000 000 002 968 590 704 988 × 2 = 0 + 0.000 000 005 937 181 409 976;
  • 4) 0.000 000 005 937 181 409 976 × 2 = 0 + 0.000 000 011 874 362 819 952;
  • 5) 0.000 000 011 874 362 819 952 × 2 = 0 + 0.000 000 023 748 725 639 904;
  • 6) 0.000 000 023 748 725 639 904 × 2 = 0 + 0.000 000 047 497 451 279 808;
  • 7) 0.000 000 047 497 451 279 808 × 2 = 0 + 0.000 000 094 994 902 559 616;
  • 8) 0.000 000 094 994 902 559 616 × 2 = 0 + 0.000 000 189 989 805 119 232;
  • 9) 0.000 000 189 989 805 119 232 × 2 = 0 + 0.000 000 379 979 610 238 464;
  • 10) 0.000 000 379 979 610 238 464 × 2 = 0 + 0.000 000 759 959 220 476 928;
  • 11) 0.000 000 759 959 220 476 928 × 2 = 0 + 0.000 001 519 918 440 953 856;
  • 12) 0.000 001 519 918 440 953 856 × 2 = 0 + 0.000 003 039 836 881 907 712;
  • 13) 0.000 003 039 836 881 907 712 × 2 = 0 + 0.000 006 079 673 763 815 424;
  • 14) 0.000 006 079 673 763 815 424 × 2 = 0 + 0.000 012 159 347 527 630 848;
  • 15) 0.000 012 159 347 527 630 848 × 2 = 0 + 0.000 024 318 695 055 261 696;
  • 16) 0.000 024 318 695 055 261 696 × 2 = 0 + 0.000 048 637 390 110 523 392;
  • 17) 0.000 048 637 390 110 523 392 × 2 = 0 + 0.000 097 274 780 221 046 784;
  • 18) 0.000 097 274 780 221 046 784 × 2 = 0 + 0.000 194 549 560 442 093 568;
  • 19) 0.000 194 549 560 442 093 568 × 2 = 0 + 0.000 389 099 120 884 187 136;
  • 20) 0.000 389 099 120 884 187 136 × 2 = 0 + 0.000 778 198 241 768 374 272;
  • 21) 0.000 778 198 241 768 374 272 × 2 = 0 + 0.001 556 396 483 536 748 544;
  • 22) 0.001 556 396 483 536 748 544 × 2 = 0 + 0.003 112 792 967 073 497 088;
  • 23) 0.003 112 792 967 073 497 088 × 2 = 0 + 0.006 225 585 934 146 994 176;
  • 24) 0.006 225 585 934 146 994 176 × 2 = 0 + 0.012 451 171 868 293 988 352;
  • 25) 0.012 451 171 868 293 988 352 × 2 = 0 + 0.024 902 343 736 587 976 704;
  • 26) 0.024 902 343 736 587 976 704 × 2 = 0 + 0.049 804 687 473 175 953 408;
  • 27) 0.049 804 687 473 175 953 408 × 2 = 0 + 0.099 609 374 946 351 906 816;
  • 28) 0.099 609 374 946 351 906 816 × 2 = 0 + 0.199 218 749 892 703 813 632;
  • 29) 0.199 218 749 892 703 813 632 × 2 = 0 + 0.398 437 499 785 407 627 264;
  • 30) 0.398 437 499 785 407 627 264 × 2 = 0 + 0.796 874 999 570 815 254 528;
  • 31) 0.796 874 999 570 815 254 528 × 2 = 1 + 0.593 749 999 141 630 509 056;
  • 32) 0.593 749 999 141 630 509 056 × 2 = 1 + 0.187 499 998 283 261 018 112;
  • 33) 0.187 499 998 283 261 018 112 × 2 = 0 + 0.374 999 996 566 522 036 224;
  • 34) 0.374 999 996 566 522 036 224 × 2 = 0 + 0.749 999 993 133 044 072 448;
  • 35) 0.749 999 993 133 044 072 448 × 2 = 1 + 0.499 999 986 266 088 144 896;
  • 36) 0.499 999 986 266 088 144 896 × 2 = 0 + 0.999 999 972 532 176 289 792;
  • 37) 0.999 999 972 532 176 289 792 × 2 = 1 + 0.999 999 945 064 352 579 584;
  • 38) 0.999 999 945 064 352 579 584 × 2 = 1 + 0.999 999 890 128 705 159 168;
  • 39) 0.999 999 890 128 705 159 168 × 2 = 1 + 0.999 999 780 257 410 318 336;
  • 40) 0.999 999 780 257 410 318 336 × 2 = 1 + 0.999 999 560 514 820 636 672;
  • 41) 0.999 999 560 514 820 636 672 × 2 = 1 + 0.999 999 121 029 641 273 344;
  • 42) 0.999 999 121 029 641 273 344 × 2 = 1 + 0.999 998 242 059 282 546 688;
  • 43) 0.999 998 242 059 282 546 688 × 2 = 1 + 0.999 996 484 118 565 093 376;
  • 44) 0.999 996 484 118 565 093 376 × 2 = 1 + 0.999 992 968 237 130 186 752;
  • 45) 0.999 992 968 237 130 186 752 × 2 = 1 + 0.999 985 936 474 260 373 504;
  • 46) 0.999 985 936 474 260 373 504 × 2 = 1 + 0.999 971 872 948 520 747 008;
  • 47) 0.999 971 872 948 520 747 008 × 2 = 1 + 0.999 943 745 897 041 494 016;
  • 48) 0.999 943 745 897 041 494 016 × 2 = 1 + 0.999 887 491 794 082 988 032;
  • 49) 0.999 887 491 794 082 988 032 × 2 = 1 + 0.999 774 983 588 165 976 064;
  • 50) 0.999 774 983 588 165 976 064 × 2 = 1 + 0.999 549 967 176 331 952 128;
  • 51) 0.999 549 967 176 331 952 128 × 2 = 1 + 0.999 099 934 352 663 904 256;
  • 52) 0.999 099 934 352 663 904 256 × 2 = 1 + 0.998 199 868 705 327 808 512;
  • 53) 0.998 199 868 705 327 808 512 × 2 = 1 + 0.996 399 737 410 655 617 024;
  • 54) 0.996 399 737 410 655 617 024 × 2 = 1 + 0.992 799 474 821 311 234 048;

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:


0.000 000 000 742 147 676 247(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 247(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:


0.000 000 000 742 147 676 247(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 -0.000 000 000 742 147 676 247 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 245 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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