-0.000 000 000 742 147 676 646 807 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 807(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
-0.000 000 000 742 147 676 646 807(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:

|-0.000 000 000 742 147 676 646 807| = 0.000 000 000 742 147 676 646 807


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: 0.000 000 000 742 147 676 646 807.

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) 0.000 000 000 742 147 676 646 807 × 2 = 0 + 0.000 000 001 484 295 353 293 614;
  • 2) 0.000 000 001 484 295 353 293 614 × 2 = 0 + 0.000 000 002 968 590 706 587 228;
  • 3) 0.000 000 002 968 590 706 587 228 × 2 = 0 + 0.000 000 005 937 181 413 174 456;
  • 4) 0.000 000 005 937 181 413 174 456 × 2 = 0 + 0.000 000 011 874 362 826 348 912;
  • 5) 0.000 000 011 874 362 826 348 912 × 2 = 0 + 0.000 000 023 748 725 652 697 824;
  • 6) 0.000 000 023 748 725 652 697 824 × 2 = 0 + 0.000 000 047 497 451 305 395 648;
  • 7) 0.000 000 047 497 451 305 395 648 × 2 = 0 + 0.000 000 094 994 902 610 791 296;
  • 8) 0.000 000 094 994 902 610 791 296 × 2 = 0 + 0.000 000 189 989 805 221 582 592;
  • 9) 0.000 000 189 989 805 221 582 592 × 2 = 0 + 0.000 000 379 979 610 443 165 184;
  • 10) 0.000 000 379 979 610 443 165 184 × 2 = 0 + 0.000 000 759 959 220 886 330 368;
  • 11) 0.000 000 759 959 220 886 330 368 × 2 = 0 + 0.000 001 519 918 441 772 660 736;
  • 12) 0.000 001 519 918 441 772 660 736 × 2 = 0 + 0.000 003 039 836 883 545 321 472;
  • 13) 0.000 003 039 836 883 545 321 472 × 2 = 0 + 0.000 006 079 673 767 090 642 944;
  • 14) 0.000 006 079 673 767 090 642 944 × 2 = 0 + 0.000 012 159 347 534 181 285 888;
  • 15) 0.000 012 159 347 534 181 285 888 × 2 = 0 + 0.000 024 318 695 068 362 571 776;
  • 16) 0.000 024 318 695 068 362 571 776 × 2 = 0 + 0.000 048 637 390 136 725 143 552;
  • 17) 0.000 048 637 390 136 725 143 552 × 2 = 0 + 0.000 097 274 780 273 450 287 104;
  • 18) 0.000 097 274 780 273 450 287 104 × 2 = 0 + 0.000 194 549 560 546 900 574 208;
  • 19) 0.000 194 549 560 546 900 574 208 × 2 = 0 + 0.000 389 099 121 093 801 148 416;
  • 20) 0.000 389 099 121 093 801 148 416 × 2 = 0 + 0.000 778 198 242 187 602 296 832;
  • 21) 0.000 778 198 242 187 602 296 832 × 2 = 0 + 0.001 556 396 484 375 204 593 664;
  • 22) 0.001 556 396 484 375 204 593 664 × 2 = 0 + 0.003 112 792 968 750 409 187 328;
  • 23) 0.003 112 792 968 750 409 187 328 × 2 = 0 + 0.006 225 585 937 500 818 374 656;
  • 24) 0.006 225 585 937 500 818 374 656 × 2 = 0 + 0.012 451 171 875 001 636 749 312;
  • 25) 0.012 451 171 875 001 636 749 312 × 2 = 0 + 0.024 902 343 750 003 273 498 624;
  • 26) 0.024 902 343 750 003 273 498 624 × 2 = 0 + 0.049 804 687 500 006 546 997 248;
  • 27) 0.049 804 687 500 006 546 997 248 × 2 = 0 + 0.099 609 375 000 013 093 994 496;
  • 28) 0.099 609 375 000 013 093 994 496 × 2 = 0 + 0.199 218 750 000 026 187 988 992;
  • 29) 0.199 218 750 000 026 187 988 992 × 2 = 0 + 0.398 437 500 000 052 375 977 984;
  • 30) 0.398 437 500 000 052 375 977 984 × 2 = 0 + 0.796 875 000 000 104 751 955 968;
  • 31) 0.796 875 000 000 104 751 955 968 × 2 = 1 + 0.593 750 000 000 209 503 911 936;
  • 32) 0.593 750 000 000 209 503 911 936 × 2 = 1 + 0.187 500 000 000 419 007 823 872;
  • 33) 0.187 500 000 000 419 007 823 872 × 2 = 0 + 0.375 000 000 000 838 015 647 744;
  • 34) 0.375 000 000 000 838 015 647 744 × 2 = 0 + 0.750 000 000 001 676 031 295 488;
  • 35) 0.750 000 000 001 676 031 295 488 × 2 = 1 + 0.500 000 000 003 352 062 590 976;
  • 36) 0.500 000 000 003 352 062 590 976 × 2 = 1 + 0.000 000 000 006 704 125 181 952;
  • 37) 0.000 000 000 006 704 125 181 952 × 2 = 0 + 0.000 000 000 013 408 250 363 904;
  • 38) 0.000 000 000 013 408 250 363 904 × 2 = 0 + 0.000 000 000 026 816 500 727 808;
  • 39) 0.000 000 000 026 816 500 727 808 × 2 = 0 + 0.000 000 000 053 633 001 455 616;
  • 40) 0.000 000 000 053 633 001 455 616 × 2 = 0 + 0.000 000 000 107 266 002 911 232;
  • 41) 0.000 000 000 107 266 002 911 232 × 2 = 0 + 0.000 000 000 214 532 005 822 464;
  • 42) 0.000 000 000 214 532 005 822 464 × 2 = 0 + 0.000 000 000 429 064 011 644 928;
  • 43) 0.000 000 000 429 064 011 644 928 × 2 = 0 + 0.000 000 000 858 128 023 289 856;
  • 44) 0.000 000 000 858 128 023 289 856 × 2 = 0 + 0.000 000 001 716 256 046 579 712;
  • 45) 0.000 000 001 716 256 046 579 712 × 2 = 0 + 0.000 000 003 432 512 093 159 424;
  • 46) 0.000 000 003 432 512 093 159 424 × 2 = 0 + 0.000 000 006 865 024 186 318 848;
  • 47) 0.000 000 006 865 024 186 318 848 × 2 = 0 + 0.000 000 013 730 048 372 637 696;
  • 48) 0.000 000 013 730 048 372 637 696 × 2 = 0 + 0.000 000 027 460 096 745 275 392;
  • 49) 0.000 000 027 460 096 745 275 392 × 2 = 0 + 0.000 000 054 920 193 490 550 784;
  • 50) 0.000 000 054 920 193 490 550 784 × 2 = 0 + 0.000 000 109 840 386 981 101 568;
  • 51) 0.000 000 109 840 386 981 101 568 × 2 = 0 + 0.000 000 219 680 773 962 203 136;
  • 52) 0.000 000 219 680 773 962 203 136 × 2 = 0 + 0.000 000 439 361 547 924 406 272;
  • 53) 0.000 000 439 361 547 924 406 272 × 2 = 0 + 0.000 000 878 723 095 848 812 544;
  • 54) 0.000 000 878 723 095 848 812 544 × 2 = 0 + 0.000 001 757 446 191 697 625 088;

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:


0.000 000 000 742 147 676 646 807(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 807(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:


0.000 000 000 742 147 676 646 807(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 -0.000 000 000 742 147 676 646 807 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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