-0.000 000 000 742 147 676 646 708 33 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 708 33(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
-0.000 000 000 742 147 676 646 708 33(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:

|-0.000 000 000 742 147 676 646 708 33| = 0.000 000 000 742 147 676 646 708 33


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: 0.000 000 000 742 147 676 646 708 33.

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) 0.000 000 000 742 147 676 646 708 33 × 2 = 0 + 0.000 000 001 484 295 353 293 416 66;
  • 2) 0.000 000 001 484 295 353 293 416 66 × 2 = 0 + 0.000 000 002 968 590 706 586 833 32;
  • 3) 0.000 000 002 968 590 706 586 833 32 × 2 = 0 + 0.000 000 005 937 181 413 173 666 64;
  • 4) 0.000 000 005 937 181 413 173 666 64 × 2 = 0 + 0.000 000 011 874 362 826 347 333 28;
  • 5) 0.000 000 011 874 362 826 347 333 28 × 2 = 0 + 0.000 000 023 748 725 652 694 666 56;
  • 6) 0.000 000 023 748 725 652 694 666 56 × 2 = 0 + 0.000 000 047 497 451 305 389 333 12;
  • 7) 0.000 000 047 497 451 305 389 333 12 × 2 = 0 + 0.000 000 094 994 902 610 778 666 24;
  • 8) 0.000 000 094 994 902 610 778 666 24 × 2 = 0 + 0.000 000 189 989 805 221 557 332 48;
  • 9) 0.000 000 189 989 805 221 557 332 48 × 2 = 0 + 0.000 000 379 979 610 443 114 664 96;
  • 10) 0.000 000 379 979 610 443 114 664 96 × 2 = 0 + 0.000 000 759 959 220 886 229 329 92;
  • 11) 0.000 000 759 959 220 886 229 329 92 × 2 = 0 + 0.000 001 519 918 441 772 458 659 84;
  • 12) 0.000 001 519 918 441 772 458 659 84 × 2 = 0 + 0.000 003 039 836 883 544 917 319 68;
  • 13) 0.000 003 039 836 883 544 917 319 68 × 2 = 0 + 0.000 006 079 673 767 089 834 639 36;
  • 14) 0.000 006 079 673 767 089 834 639 36 × 2 = 0 + 0.000 012 159 347 534 179 669 278 72;
  • 15) 0.000 012 159 347 534 179 669 278 72 × 2 = 0 + 0.000 024 318 695 068 359 338 557 44;
  • 16) 0.000 024 318 695 068 359 338 557 44 × 2 = 0 + 0.000 048 637 390 136 718 677 114 88;
  • 17) 0.000 048 637 390 136 718 677 114 88 × 2 = 0 + 0.000 097 274 780 273 437 354 229 76;
  • 18) 0.000 097 274 780 273 437 354 229 76 × 2 = 0 + 0.000 194 549 560 546 874 708 459 52;
  • 19) 0.000 194 549 560 546 874 708 459 52 × 2 = 0 + 0.000 389 099 121 093 749 416 919 04;
  • 20) 0.000 389 099 121 093 749 416 919 04 × 2 = 0 + 0.000 778 198 242 187 498 833 838 08;
  • 21) 0.000 778 198 242 187 498 833 838 08 × 2 = 0 + 0.001 556 396 484 374 997 667 676 16;
  • 22) 0.001 556 396 484 374 997 667 676 16 × 2 = 0 + 0.003 112 792 968 749 995 335 352 32;
  • 23) 0.003 112 792 968 749 995 335 352 32 × 2 = 0 + 0.006 225 585 937 499 990 670 704 64;
  • 24) 0.006 225 585 937 499 990 670 704 64 × 2 = 0 + 0.012 451 171 874 999 981 341 409 28;
  • 25) 0.012 451 171 874 999 981 341 409 28 × 2 = 0 + 0.024 902 343 749 999 962 682 818 56;
  • 26) 0.024 902 343 749 999 962 682 818 56 × 2 = 0 + 0.049 804 687 499 999 925 365 637 12;
  • 27) 0.049 804 687 499 999 925 365 637 12 × 2 = 0 + 0.099 609 374 999 999 850 731 274 24;
  • 28) 0.099 609 374 999 999 850 731 274 24 × 2 = 0 + 0.199 218 749 999 999 701 462 548 48;
  • 29) 0.199 218 749 999 999 701 462 548 48 × 2 = 0 + 0.398 437 499 999 999 402 925 096 96;
  • 30) 0.398 437 499 999 999 402 925 096 96 × 2 = 0 + 0.796 874 999 999 998 805 850 193 92;
  • 31) 0.796 874 999 999 998 805 850 193 92 × 2 = 1 + 0.593 749 999 999 997 611 700 387 84;
  • 32) 0.593 749 999 999 997 611 700 387 84 × 2 = 1 + 0.187 499 999 999 995 223 400 775 68;
  • 33) 0.187 499 999 999 995 223 400 775 68 × 2 = 0 + 0.374 999 999 999 990 446 801 551 36;
  • 34) 0.374 999 999 999 990 446 801 551 36 × 2 = 0 + 0.749 999 999 999 980 893 603 102 72;
  • 35) 0.749 999 999 999 980 893 603 102 72 × 2 = 1 + 0.499 999 999 999 961 787 206 205 44;
  • 36) 0.499 999 999 999 961 787 206 205 44 × 2 = 0 + 0.999 999 999 999 923 574 412 410 88;
  • 37) 0.999 999 999 999 923 574 412 410 88 × 2 = 1 + 0.999 999 999 999 847 148 824 821 76;
  • 38) 0.999 999 999 999 847 148 824 821 76 × 2 = 1 + 0.999 999 999 999 694 297 649 643 52;
  • 39) 0.999 999 999 999 694 297 649 643 52 × 2 = 1 + 0.999 999 999 999 388 595 299 287 04;
  • 40) 0.999 999 999 999 388 595 299 287 04 × 2 = 1 + 0.999 999 999 998 777 190 598 574 08;
  • 41) 0.999 999 999 998 777 190 598 574 08 × 2 = 1 + 0.999 999 999 997 554 381 197 148 16;
  • 42) 0.999 999 999 997 554 381 197 148 16 × 2 = 1 + 0.999 999 999 995 108 762 394 296 32;
  • 43) 0.999 999 999 995 108 762 394 296 32 × 2 = 1 + 0.999 999 999 990 217 524 788 592 64;
  • 44) 0.999 999 999 990 217 524 788 592 64 × 2 = 1 + 0.999 999 999 980 435 049 577 185 28;
  • 45) 0.999 999 999 980 435 049 577 185 28 × 2 = 1 + 0.999 999 999 960 870 099 154 370 56;
  • 46) 0.999 999 999 960 870 099 154 370 56 × 2 = 1 + 0.999 999 999 921 740 198 308 741 12;
  • 47) 0.999 999 999 921 740 198 308 741 12 × 2 = 1 + 0.999 999 999 843 480 396 617 482 24;
  • 48) 0.999 999 999 843 480 396 617 482 24 × 2 = 1 + 0.999 999 999 686 960 793 234 964 48;
  • 49) 0.999 999 999 686 960 793 234 964 48 × 2 = 1 + 0.999 999 999 373 921 586 469 928 96;
  • 50) 0.999 999 999 373 921 586 469 928 96 × 2 = 1 + 0.999 999 998 747 843 172 939 857 92;
  • 51) 0.999 999 998 747 843 172 939 857 92 × 2 = 1 + 0.999 999 997 495 686 345 879 715 84;
  • 52) 0.999 999 997 495 686 345 879 715 84 × 2 = 1 + 0.999 999 994 991 372 691 759 431 68;
  • 53) 0.999 999 994 991 372 691 759 431 68 × 2 = 1 + 0.999 999 989 982 745 383 518 863 36;
  • 54) 0.999 999 989 982 745 383 518 863 36 × 2 = 1 + 0.999 999 979 965 490 767 037 726 72;

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:


0.000 000 000 742 147 676 646 708 33(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 708 33(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:


0.000 000 000 742 147 676 646 708 33(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 -0.000 000 000 742 147 676 646 708 33 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