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

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

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


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: 0.000 000 000 742 147 676 646 742.

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) 0.000 000 000 742 147 676 646 742 × 2 = 0 + 0.000 000 001 484 295 353 293 484;
  • 2) 0.000 000 001 484 295 353 293 484 × 2 = 0 + 0.000 000 002 968 590 706 586 968;
  • 3) 0.000 000 002 968 590 706 586 968 × 2 = 0 + 0.000 000 005 937 181 413 173 936;
  • 4) 0.000 000 005 937 181 413 173 936 × 2 = 0 + 0.000 000 011 874 362 826 347 872;
  • 5) 0.000 000 011 874 362 826 347 872 × 2 = 0 + 0.000 000 023 748 725 652 695 744;
  • 6) 0.000 000 023 748 725 652 695 744 × 2 = 0 + 0.000 000 047 497 451 305 391 488;
  • 7) 0.000 000 047 497 451 305 391 488 × 2 = 0 + 0.000 000 094 994 902 610 782 976;
  • 8) 0.000 000 094 994 902 610 782 976 × 2 = 0 + 0.000 000 189 989 805 221 565 952;
  • 9) 0.000 000 189 989 805 221 565 952 × 2 = 0 + 0.000 000 379 979 610 443 131 904;
  • 10) 0.000 000 379 979 610 443 131 904 × 2 = 0 + 0.000 000 759 959 220 886 263 808;
  • 11) 0.000 000 759 959 220 886 263 808 × 2 = 0 + 0.000 001 519 918 441 772 527 616;
  • 12) 0.000 001 519 918 441 772 527 616 × 2 = 0 + 0.000 003 039 836 883 545 055 232;
  • 13) 0.000 003 039 836 883 545 055 232 × 2 = 0 + 0.000 006 079 673 767 090 110 464;
  • 14) 0.000 006 079 673 767 090 110 464 × 2 = 0 + 0.000 012 159 347 534 180 220 928;
  • 15) 0.000 012 159 347 534 180 220 928 × 2 = 0 + 0.000 024 318 695 068 360 441 856;
  • 16) 0.000 024 318 695 068 360 441 856 × 2 = 0 + 0.000 048 637 390 136 720 883 712;
  • 17) 0.000 048 637 390 136 720 883 712 × 2 = 0 + 0.000 097 274 780 273 441 767 424;
  • 18) 0.000 097 274 780 273 441 767 424 × 2 = 0 + 0.000 194 549 560 546 883 534 848;
  • 19) 0.000 194 549 560 546 883 534 848 × 2 = 0 + 0.000 389 099 121 093 767 069 696;
  • 20) 0.000 389 099 121 093 767 069 696 × 2 = 0 + 0.000 778 198 242 187 534 139 392;
  • 21) 0.000 778 198 242 187 534 139 392 × 2 = 0 + 0.001 556 396 484 375 068 278 784;
  • 22) 0.001 556 396 484 375 068 278 784 × 2 = 0 + 0.003 112 792 968 750 136 557 568;
  • 23) 0.003 112 792 968 750 136 557 568 × 2 = 0 + 0.006 225 585 937 500 273 115 136;
  • 24) 0.006 225 585 937 500 273 115 136 × 2 = 0 + 0.012 451 171 875 000 546 230 272;
  • 25) 0.012 451 171 875 000 546 230 272 × 2 = 0 + 0.024 902 343 750 001 092 460 544;
  • 26) 0.024 902 343 750 001 092 460 544 × 2 = 0 + 0.049 804 687 500 002 184 921 088;
  • 27) 0.049 804 687 500 002 184 921 088 × 2 = 0 + 0.099 609 375 000 004 369 842 176;
  • 28) 0.099 609 375 000 004 369 842 176 × 2 = 0 + 0.199 218 750 000 008 739 684 352;
  • 29) 0.199 218 750 000 008 739 684 352 × 2 = 0 + 0.398 437 500 000 017 479 368 704;
  • 30) 0.398 437 500 000 017 479 368 704 × 2 = 0 + 0.796 875 000 000 034 958 737 408;
  • 31) 0.796 875 000 000 034 958 737 408 × 2 = 1 + 0.593 750 000 000 069 917 474 816;
  • 32) 0.593 750 000 000 069 917 474 816 × 2 = 1 + 0.187 500 000 000 139 834 949 632;
  • 33) 0.187 500 000 000 139 834 949 632 × 2 = 0 + 0.375 000 000 000 279 669 899 264;
  • 34) 0.375 000 000 000 279 669 899 264 × 2 = 0 + 0.750 000 000 000 559 339 798 528;
  • 35) 0.750 000 000 000 559 339 798 528 × 2 = 1 + 0.500 000 000 001 118 679 597 056;
  • 36) 0.500 000 000 001 118 679 597 056 × 2 = 1 + 0.000 000 000 002 237 359 194 112;
  • 37) 0.000 000 000 002 237 359 194 112 × 2 = 0 + 0.000 000 000 004 474 718 388 224;
  • 38) 0.000 000 000 004 474 718 388 224 × 2 = 0 + 0.000 000 000 008 949 436 776 448;
  • 39) 0.000 000 000 008 949 436 776 448 × 2 = 0 + 0.000 000 000 017 898 873 552 896;
  • 40) 0.000 000 000 017 898 873 552 896 × 2 = 0 + 0.000 000 000 035 797 747 105 792;
  • 41) 0.000 000 000 035 797 747 105 792 × 2 = 0 + 0.000 000 000 071 595 494 211 584;
  • 42) 0.000 000 000 071 595 494 211 584 × 2 = 0 + 0.000 000 000 143 190 988 423 168;
  • 43) 0.000 000 000 143 190 988 423 168 × 2 = 0 + 0.000 000 000 286 381 976 846 336;
  • 44) 0.000 000 000 286 381 976 846 336 × 2 = 0 + 0.000 000 000 572 763 953 692 672;
  • 45) 0.000 000 000 572 763 953 692 672 × 2 = 0 + 0.000 000 001 145 527 907 385 344;
  • 46) 0.000 000 001 145 527 907 385 344 × 2 = 0 + 0.000 000 002 291 055 814 770 688;
  • 47) 0.000 000 002 291 055 814 770 688 × 2 = 0 + 0.000 000 004 582 111 629 541 376;
  • 48) 0.000 000 004 582 111 629 541 376 × 2 = 0 + 0.000 000 009 164 223 259 082 752;
  • 49) 0.000 000 009 164 223 259 082 752 × 2 = 0 + 0.000 000 018 328 446 518 165 504;
  • 50) 0.000 000 018 328 446 518 165 504 × 2 = 0 + 0.000 000 036 656 893 036 331 008;
  • 51) 0.000 000 036 656 893 036 331 008 × 2 = 0 + 0.000 000 073 313 786 072 662 016;
  • 52) 0.000 000 073 313 786 072 662 016 × 2 = 0 + 0.000 000 146 627 572 145 324 032;
  • 53) 0.000 000 146 627 572 145 324 032 × 2 = 0 + 0.000 000 293 255 144 290 648 064;
  • 54) 0.000 000 293 255 144 290 648 064 × 2 = 0 + 0.000 000 586 510 288 581 296 128;

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:


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

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 742(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:


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

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 646 756 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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