-0.000 000 000 742 147 676 646 667 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 667(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
-0.000 000 000 742 147 676 646 667(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:

|-0.000 000 000 742 147 676 646 667| = 0.000 000 000 742 147 676 646 667


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: 0.000 000 000 742 147 676 646 667.

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) 0.000 000 000 742 147 676 646 667 × 2 = 0 + 0.000 000 001 484 295 353 293 334;
  • 2) 0.000 000 001 484 295 353 293 334 × 2 = 0 + 0.000 000 002 968 590 706 586 668;
  • 3) 0.000 000 002 968 590 706 586 668 × 2 = 0 + 0.000 000 005 937 181 413 173 336;
  • 4) 0.000 000 005 937 181 413 173 336 × 2 = 0 + 0.000 000 011 874 362 826 346 672;
  • 5) 0.000 000 011 874 362 826 346 672 × 2 = 0 + 0.000 000 023 748 725 652 693 344;
  • 6) 0.000 000 023 748 725 652 693 344 × 2 = 0 + 0.000 000 047 497 451 305 386 688;
  • 7) 0.000 000 047 497 451 305 386 688 × 2 = 0 + 0.000 000 094 994 902 610 773 376;
  • 8) 0.000 000 094 994 902 610 773 376 × 2 = 0 + 0.000 000 189 989 805 221 546 752;
  • 9) 0.000 000 189 989 805 221 546 752 × 2 = 0 + 0.000 000 379 979 610 443 093 504;
  • 10) 0.000 000 379 979 610 443 093 504 × 2 = 0 + 0.000 000 759 959 220 886 187 008;
  • 11) 0.000 000 759 959 220 886 187 008 × 2 = 0 + 0.000 001 519 918 441 772 374 016;
  • 12) 0.000 001 519 918 441 772 374 016 × 2 = 0 + 0.000 003 039 836 883 544 748 032;
  • 13) 0.000 003 039 836 883 544 748 032 × 2 = 0 + 0.000 006 079 673 767 089 496 064;
  • 14) 0.000 006 079 673 767 089 496 064 × 2 = 0 + 0.000 012 159 347 534 178 992 128;
  • 15) 0.000 012 159 347 534 178 992 128 × 2 = 0 + 0.000 024 318 695 068 357 984 256;
  • 16) 0.000 024 318 695 068 357 984 256 × 2 = 0 + 0.000 048 637 390 136 715 968 512;
  • 17) 0.000 048 637 390 136 715 968 512 × 2 = 0 + 0.000 097 274 780 273 431 937 024;
  • 18) 0.000 097 274 780 273 431 937 024 × 2 = 0 + 0.000 194 549 560 546 863 874 048;
  • 19) 0.000 194 549 560 546 863 874 048 × 2 = 0 + 0.000 389 099 121 093 727 748 096;
  • 20) 0.000 389 099 121 093 727 748 096 × 2 = 0 + 0.000 778 198 242 187 455 496 192;
  • 21) 0.000 778 198 242 187 455 496 192 × 2 = 0 + 0.001 556 396 484 374 910 992 384;
  • 22) 0.001 556 396 484 374 910 992 384 × 2 = 0 + 0.003 112 792 968 749 821 984 768;
  • 23) 0.003 112 792 968 749 821 984 768 × 2 = 0 + 0.006 225 585 937 499 643 969 536;
  • 24) 0.006 225 585 937 499 643 969 536 × 2 = 0 + 0.012 451 171 874 999 287 939 072;
  • 25) 0.012 451 171 874 999 287 939 072 × 2 = 0 + 0.024 902 343 749 998 575 878 144;
  • 26) 0.024 902 343 749 998 575 878 144 × 2 = 0 + 0.049 804 687 499 997 151 756 288;
  • 27) 0.049 804 687 499 997 151 756 288 × 2 = 0 + 0.099 609 374 999 994 303 512 576;
  • 28) 0.099 609 374 999 994 303 512 576 × 2 = 0 + 0.199 218 749 999 988 607 025 152;
  • 29) 0.199 218 749 999 988 607 025 152 × 2 = 0 + 0.398 437 499 999 977 214 050 304;
  • 30) 0.398 437 499 999 977 214 050 304 × 2 = 0 + 0.796 874 999 999 954 428 100 608;
  • 31) 0.796 874 999 999 954 428 100 608 × 2 = 1 + 0.593 749 999 999 908 856 201 216;
  • 32) 0.593 749 999 999 908 856 201 216 × 2 = 1 + 0.187 499 999 999 817 712 402 432;
  • 33) 0.187 499 999 999 817 712 402 432 × 2 = 0 + 0.374 999 999 999 635 424 804 864;
  • 34) 0.374 999 999 999 635 424 804 864 × 2 = 0 + 0.749 999 999 999 270 849 609 728;
  • 35) 0.749 999 999 999 270 849 609 728 × 2 = 1 + 0.499 999 999 998 541 699 219 456;
  • 36) 0.499 999 999 998 541 699 219 456 × 2 = 0 + 0.999 999 999 997 083 398 438 912;
  • 37) 0.999 999 999 997 083 398 438 912 × 2 = 1 + 0.999 999 999 994 166 796 877 824;
  • 38) 0.999 999 999 994 166 796 877 824 × 2 = 1 + 0.999 999 999 988 333 593 755 648;
  • 39) 0.999 999 999 988 333 593 755 648 × 2 = 1 + 0.999 999 999 976 667 187 511 296;
  • 40) 0.999 999 999 976 667 187 511 296 × 2 = 1 + 0.999 999 999 953 334 375 022 592;
  • 41) 0.999 999 999 953 334 375 022 592 × 2 = 1 + 0.999 999 999 906 668 750 045 184;
  • 42) 0.999 999 999 906 668 750 045 184 × 2 = 1 + 0.999 999 999 813 337 500 090 368;
  • 43) 0.999 999 999 813 337 500 090 368 × 2 = 1 + 0.999 999 999 626 675 000 180 736;
  • 44) 0.999 999 999 626 675 000 180 736 × 2 = 1 + 0.999 999 999 253 350 000 361 472;
  • 45) 0.999 999 999 253 350 000 361 472 × 2 = 1 + 0.999 999 998 506 700 000 722 944;
  • 46) 0.999 999 998 506 700 000 722 944 × 2 = 1 + 0.999 999 997 013 400 001 445 888;
  • 47) 0.999 999 997 013 400 001 445 888 × 2 = 1 + 0.999 999 994 026 800 002 891 776;
  • 48) 0.999 999 994 026 800 002 891 776 × 2 = 1 + 0.999 999 988 053 600 005 783 552;
  • 49) 0.999 999 988 053 600 005 783 552 × 2 = 1 + 0.999 999 976 107 200 011 567 104;
  • 50) 0.999 999 976 107 200 011 567 104 × 2 = 1 + 0.999 999 952 214 400 023 134 208;
  • 51) 0.999 999 952 214 400 023 134 208 × 2 = 1 + 0.999 999 904 428 800 046 268 416;
  • 52) 0.999 999 904 428 800 046 268 416 × 2 = 1 + 0.999 999 808 857 600 092 536 832;
  • 53) 0.999 999 808 857 600 092 536 832 × 2 = 1 + 0.999 999 617 715 200 185 073 664;
  • 54) 0.999 999 617 715 200 185 073 664 × 2 = 1 + 0.999 999 235 430 400 370 147 328;

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:


0.000 000 000 742 147 676 646 667(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 667(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:


0.000 000 000 742 147 676 646 667(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 -0.000 000 000 742 147 676 646 667 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