-0.000 000 000 742 147 676 061 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 061(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
-0.000 000 000 742 147 676 061(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:

|-0.000 000 000 742 147 676 061| = 0.000 000 000 742 147 676 061


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: 0.000 000 000 742 147 676 061.

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) 0.000 000 000 742 147 676 061 × 2 = 0 + 0.000 000 001 484 295 352 122;
  • 2) 0.000 000 001 484 295 352 122 × 2 = 0 + 0.000 000 002 968 590 704 244;
  • 3) 0.000 000 002 968 590 704 244 × 2 = 0 + 0.000 000 005 937 181 408 488;
  • 4) 0.000 000 005 937 181 408 488 × 2 = 0 + 0.000 000 011 874 362 816 976;
  • 5) 0.000 000 011 874 362 816 976 × 2 = 0 + 0.000 000 023 748 725 633 952;
  • 6) 0.000 000 023 748 725 633 952 × 2 = 0 + 0.000 000 047 497 451 267 904;
  • 7) 0.000 000 047 497 451 267 904 × 2 = 0 + 0.000 000 094 994 902 535 808;
  • 8) 0.000 000 094 994 902 535 808 × 2 = 0 + 0.000 000 189 989 805 071 616;
  • 9) 0.000 000 189 989 805 071 616 × 2 = 0 + 0.000 000 379 979 610 143 232;
  • 10) 0.000 000 379 979 610 143 232 × 2 = 0 + 0.000 000 759 959 220 286 464;
  • 11) 0.000 000 759 959 220 286 464 × 2 = 0 + 0.000 001 519 918 440 572 928;
  • 12) 0.000 001 519 918 440 572 928 × 2 = 0 + 0.000 003 039 836 881 145 856;
  • 13) 0.000 003 039 836 881 145 856 × 2 = 0 + 0.000 006 079 673 762 291 712;
  • 14) 0.000 006 079 673 762 291 712 × 2 = 0 + 0.000 012 159 347 524 583 424;
  • 15) 0.000 012 159 347 524 583 424 × 2 = 0 + 0.000 024 318 695 049 166 848;
  • 16) 0.000 024 318 695 049 166 848 × 2 = 0 + 0.000 048 637 390 098 333 696;
  • 17) 0.000 048 637 390 098 333 696 × 2 = 0 + 0.000 097 274 780 196 667 392;
  • 18) 0.000 097 274 780 196 667 392 × 2 = 0 + 0.000 194 549 560 393 334 784;
  • 19) 0.000 194 549 560 393 334 784 × 2 = 0 + 0.000 389 099 120 786 669 568;
  • 20) 0.000 389 099 120 786 669 568 × 2 = 0 + 0.000 778 198 241 573 339 136;
  • 21) 0.000 778 198 241 573 339 136 × 2 = 0 + 0.001 556 396 483 146 678 272;
  • 22) 0.001 556 396 483 146 678 272 × 2 = 0 + 0.003 112 792 966 293 356 544;
  • 23) 0.003 112 792 966 293 356 544 × 2 = 0 + 0.006 225 585 932 586 713 088;
  • 24) 0.006 225 585 932 586 713 088 × 2 = 0 + 0.012 451 171 865 173 426 176;
  • 25) 0.012 451 171 865 173 426 176 × 2 = 0 + 0.024 902 343 730 346 852 352;
  • 26) 0.024 902 343 730 346 852 352 × 2 = 0 + 0.049 804 687 460 693 704 704;
  • 27) 0.049 804 687 460 693 704 704 × 2 = 0 + 0.099 609 374 921 387 409 408;
  • 28) 0.099 609 374 921 387 409 408 × 2 = 0 + 0.199 218 749 842 774 818 816;
  • 29) 0.199 218 749 842 774 818 816 × 2 = 0 + 0.398 437 499 685 549 637 632;
  • 30) 0.398 437 499 685 549 637 632 × 2 = 0 + 0.796 874 999 371 099 275 264;
  • 31) 0.796 874 999 371 099 275 264 × 2 = 1 + 0.593 749 998 742 198 550 528;
  • 32) 0.593 749 998 742 198 550 528 × 2 = 1 + 0.187 499 997 484 397 101 056;
  • 33) 0.187 499 997 484 397 101 056 × 2 = 0 + 0.374 999 994 968 794 202 112;
  • 34) 0.374 999 994 968 794 202 112 × 2 = 0 + 0.749 999 989 937 588 404 224;
  • 35) 0.749 999 989 937 588 404 224 × 2 = 1 + 0.499 999 979 875 176 808 448;
  • 36) 0.499 999 979 875 176 808 448 × 2 = 0 + 0.999 999 959 750 353 616 896;
  • 37) 0.999 999 959 750 353 616 896 × 2 = 1 + 0.999 999 919 500 707 233 792;
  • 38) 0.999 999 919 500 707 233 792 × 2 = 1 + 0.999 999 839 001 414 467 584;
  • 39) 0.999 999 839 001 414 467 584 × 2 = 1 + 0.999 999 678 002 828 935 168;
  • 40) 0.999 999 678 002 828 935 168 × 2 = 1 + 0.999 999 356 005 657 870 336;
  • 41) 0.999 999 356 005 657 870 336 × 2 = 1 + 0.999 998 712 011 315 740 672;
  • 42) 0.999 998 712 011 315 740 672 × 2 = 1 + 0.999 997 424 022 631 481 344;
  • 43) 0.999 997 424 022 631 481 344 × 2 = 1 + 0.999 994 848 045 262 962 688;
  • 44) 0.999 994 848 045 262 962 688 × 2 = 1 + 0.999 989 696 090 525 925 376;
  • 45) 0.999 989 696 090 525 925 376 × 2 = 1 + 0.999 979 392 181 051 850 752;
  • 46) 0.999 979 392 181 051 850 752 × 2 = 1 + 0.999 958 784 362 103 701 504;
  • 47) 0.999 958 784 362 103 701 504 × 2 = 1 + 0.999 917 568 724 207 403 008;
  • 48) 0.999 917 568 724 207 403 008 × 2 = 1 + 0.999 835 137 448 414 806 016;
  • 49) 0.999 835 137 448 414 806 016 × 2 = 1 + 0.999 670 274 896 829 612 032;
  • 50) 0.999 670 274 896 829 612 032 × 2 = 1 + 0.999 340 549 793 659 224 064;
  • 51) 0.999 340 549 793 659 224 064 × 2 = 1 + 0.998 681 099 587 318 448 128;
  • 52) 0.998 681 099 587 318 448 128 × 2 = 1 + 0.997 362 199 174 636 896 256;
  • 53) 0.997 362 199 174 636 896 256 × 2 = 1 + 0.994 724 398 349 273 792 512;
  • 54) 0.994 724 398 349 273 792 512 × 2 = 1 + 0.989 448 796 698 547 585 024;

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:


0.000 000 000 742 147 676 061(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 061(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:


0.000 000 000 742 147 676 061(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 -0.000 000 000 742 147 676 061 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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