-0.000 000 000 742 147 676 636 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 636 3(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
-0.000 000 000 742 147 676 636 3(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:

|-0.000 000 000 742 147 676 636 3| = 0.000 000 000 742 147 676 636 3


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: 0.000 000 000 742 147 676 636 3.

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) 0.000 000 000 742 147 676 636 3 × 2 = 0 + 0.000 000 001 484 295 353 272 6;
  • 2) 0.000 000 001 484 295 353 272 6 × 2 = 0 + 0.000 000 002 968 590 706 545 2;
  • 3) 0.000 000 002 968 590 706 545 2 × 2 = 0 + 0.000 000 005 937 181 413 090 4;
  • 4) 0.000 000 005 937 181 413 090 4 × 2 = 0 + 0.000 000 011 874 362 826 180 8;
  • 5) 0.000 000 011 874 362 826 180 8 × 2 = 0 + 0.000 000 023 748 725 652 361 6;
  • 6) 0.000 000 023 748 725 652 361 6 × 2 = 0 + 0.000 000 047 497 451 304 723 2;
  • 7) 0.000 000 047 497 451 304 723 2 × 2 = 0 + 0.000 000 094 994 902 609 446 4;
  • 8) 0.000 000 094 994 902 609 446 4 × 2 = 0 + 0.000 000 189 989 805 218 892 8;
  • 9) 0.000 000 189 989 805 218 892 8 × 2 = 0 + 0.000 000 379 979 610 437 785 6;
  • 10) 0.000 000 379 979 610 437 785 6 × 2 = 0 + 0.000 000 759 959 220 875 571 2;
  • 11) 0.000 000 759 959 220 875 571 2 × 2 = 0 + 0.000 001 519 918 441 751 142 4;
  • 12) 0.000 001 519 918 441 751 142 4 × 2 = 0 + 0.000 003 039 836 883 502 284 8;
  • 13) 0.000 003 039 836 883 502 284 8 × 2 = 0 + 0.000 006 079 673 767 004 569 6;
  • 14) 0.000 006 079 673 767 004 569 6 × 2 = 0 + 0.000 012 159 347 534 009 139 2;
  • 15) 0.000 012 159 347 534 009 139 2 × 2 = 0 + 0.000 024 318 695 068 018 278 4;
  • 16) 0.000 024 318 695 068 018 278 4 × 2 = 0 + 0.000 048 637 390 136 036 556 8;
  • 17) 0.000 048 637 390 136 036 556 8 × 2 = 0 + 0.000 097 274 780 272 073 113 6;
  • 18) 0.000 097 274 780 272 073 113 6 × 2 = 0 + 0.000 194 549 560 544 146 227 2;
  • 19) 0.000 194 549 560 544 146 227 2 × 2 = 0 + 0.000 389 099 121 088 292 454 4;
  • 20) 0.000 389 099 121 088 292 454 4 × 2 = 0 + 0.000 778 198 242 176 584 908 8;
  • 21) 0.000 778 198 242 176 584 908 8 × 2 = 0 + 0.001 556 396 484 353 169 817 6;
  • 22) 0.001 556 396 484 353 169 817 6 × 2 = 0 + 0.003 112 792 968 706 339 635 2;
  • 23) 0.003 112 792 968 706 339 635 2 × 2 = 0 + 0.006 225 585 937 412 679 270 4;
  • 24) 0.006 225 585 937 412 679 270 4 × 2 = 0 + 0.012 451 171 874 825 358 540 8;
  • 25) 0.012 451 171 874 825 358 540 8 × 2 = 0 + 0.024 902 343 749 650 717 081 6;
  • 26) 0.024 902 343 749 650 717 081 6 × 2 = 0 + 0.049 804 687 499 301 434 163 2;
  • 27) 0.049 804 687 499 301 434 163 2 × 2 = 0 + 0.099 609 374 998 602 868 326 4;
  • 28) 0.099 609 374 998 602 868 326 4 × 2 = 0 + 0.199 218 749 997 205 736 652 8;
  • 29) 0.199 218 749 997 205 736 652 8 × 2 = 0 + 0.398 437 499 994 411 473 305 6;
  • 30) 0.398 437 499 994 411 473 305 6 × 2 = 0 + 0.796 874 999 988 822 946 611 2;
  • 31) 0.796 874 999 988 822 946 611 2 × 2 = 1 + 0.593 749 999 977 645 893 222 4;
  • 32) 0.593 749 999 977 645 893 222 4 × 2 = 1 + 0.187 499 999 955 291 786 444 8;
  • 33) 0.187 499 999 955 291 786 444 8 × 2 = 0 + 0.374 999 999 910 583 572 889 6;
  • 34) 0.374 999 999 910 583 572 889 6 × 2 = 0 + 0.749 999 999 821 167 145 779 2;
  • 35) 0.749 999 999 821 167 145 779 2 × 2 = 1 + 0.499 999 999 642 334 291 558 4;
  • 36) 0.499 999 999 642 334 291 558 4 × 2 = 0 + 0.999 999 999 284 668 583 116 8;
  • 37) 0.999 999 999 284 668 583 116 8 × 2 = 1 + 0.999 999 998 569 337 166 233 6;
  • 38) 0.999 999 998 569 337 166 233 6 × 2 = 1 + 0.999 999 997 138 674 332 467 2;
  • 39) 0.999 999 997 138 674 332 467 2 × 2 = 1 + 0.999 999 994 277 348 664 934 4;
  • 40) 0.999 999 994 277 348 664 934 4 × 2 = 1 + 0.999 999 988 554 697 329 868 8;
  • 41) 0.999 999 988 554 697 329 868 8 × 2 = 1 + 0.999 999 977 109 394 659 737 6;
  • 42) 0.999 999 977 109 394 659 737 6 × 2 = 1 + 0.999 999 954 218 789 319 475 2;
  • 43) 0.999 999 954 218 789 319 475 2 × 2 = 1 + 0.999 999 908 437 578 638 950 4;
  • 44) 0.999 999 908 437 578 638 950 4 × 2 = 1 + 0.999 999 816 875 157 277 900 8;
  • 45) 0.999 999 816 875 157 277 900 8 × 2 = 1 + 0.999 999 633 750 314 555 801 6;
  • 46) 0.999 999 633 750 314 555 801 6 × 2 = 1 + 0.999 999 267 500 629 111 603 2;
  • 47) 0.999 999 267 500 629 111 603 2 × 2 = 1 + 0.999 998 535 001 258 223 206 4;
  • 48) 0.999 998 535 001 258 223 206 4 × 2 = 1 + 0.999 997 070 002 516 446 412 8;
  • 49) 0.999 997 070 002 516 446 412 8 × 2 = 1 + 0.999 994 140 005 032 892 825 6;
  • 50) 0.999 994 140 005 032 892 825 6 × 2 = 1 + 0.999 988 280 010 065 785 651 2;
  • 51) 0.999 988 280 010 065 785 651 2 × 2 = 1 + 0.999 976 560 020 131 571 302 4;
  • 52) 0.999 976 560 020 131 571 302 4 × 2 = 1 + 0.999 953 120 040 263 142 604 8;
  • 53) 0.999 953 120 040 263 142 604 8 × 2 = 1 + 0.999 906 240 080 526 285 209 6;
  • 54) 0.999 906 240 080 526 285 209 6 × 2 = 1 + 0.999 812 480 161 052 570 419 2;

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:


0.000 000 000 742 147 676 636 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 636 3(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:


0.000 000 000 742 147 676 636 3(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 -0.000 000 000 742 147 676 636 3 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