2.858 648 867 222 626 824 684 408 350 103 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.858 648 867 222 626 824 684 408 350 103(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
2.858 648 867 222 626 824 684 408 350 103(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. First, convert to binary (in base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.858 648 867 222 626 824 684 408 350 103.

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) 0.858 648 867 222 626 824 684 408 350 103 × 2 = 1 + 0.717 297 734 445 253 649 368 816 700 206;
  • 2) 0.717 297 734 445 253 649 368 816 700 206 × 2 = 1 + 0.434 595 468 890 507 298 737 633 400 412;
  • 3) 0.434 595 468 890 507 298 737 633 400 412 × 2 = 0 + 0.869 190 937 781 014 597 475 266 800 824;
  • 4) 0.869 190 937 781 014 597 475 266 800 824 × 2 = 1 + 0.738 381 875 562 029 194 950 533 601 648;
  • 5) 0.738 381 875 562 029 194 950 533 601 648 × 2 = 1 + 0.476 763 751 124 058 389 901 067 203 296;
  • 6) 0.476 763 751 124 058 389 901 067 203 296 × 2 = 0 + 0.953 527 502 248 116 779 802 134 406 592;
  • 7) 0.953 527 502 248 116 779 802 134 406 592 × 2 = 1 + 0.907 055 004 496 233 559 604 268 813 184;
  • 8) 0.907 055 004 496 233 559 604 268 813 184 × 2 = 1 + 0.814 110 008 992 467 119 208 537 626 368;
  • 9) 0.814 110 008 992 467 119 208 537 626 368 × 2 = 1 + 0.628 220 017 984 934 238 417 075 252 736;
  • 10) 0.628 220 017 984 934 238 417 075 252 736 × 2 = 1 + 0.256 440 035 969 868 476 834 150 505 472;
  • 11) 0.256 440 035 969 868 476 834 150 505 472 × 2 = 0 + 0.512 880 071 939 736 953 668 301 010 944;
  • 12) 0.512 880 071 939 736 953 668 301 010 944 × 2 = 1 + 0.025 760 143 879 473 907 336 602 021 888;
  • 13) 0.025 760 143 879 473 907 336 602 021 888 × 2 = 0 + 0.051 520 287 758 947 814 673 204 043 776;
  • 14) 0.051 520 287 758 947 814 673 204 043 776 × 2 = 0 + 0.103 040 575 517 895 629 346 408 087 552;
  • 15) 0.103 040 575 517 895 629 346 408 087 552 × 2 = 0 + 0.206 081 151 035 791 258 692 816 175 104;
  • 16) 0.206 081 151 035 791 258 692 816 175 104 × 2 = 0 + 0.412 162 302 071 582 517 385 632 350 208;
  • 17) 0.412 162 302 071 582 517 385 632 350 208 × 2 = 0 + 0.824 324 604 143 165 034 771 264 700 416;
  • 18) 0.824 324 604 143 165 034 771 264 700 416 × 2 = 1 + 0.648 649 208 286 330 069 542 529 400 832;
  • 19) 0.648 649 208 286 330 069 542 529 400 832 × 2 = 1 + 0.297 298 416 572 660 139 085 058 801 664;
  • 20) 0.297 298 416 572 660 139 085 058 801 664 × 2 = 0 + 0.594 596 833 145 320 278 170 117 603 328;
  • 21) 0.594 596 833 145 320 278 170 117 603 328 × 2 = 1 + 0.189 193 666 290 640 556 340 235 206 656;
  • 22) 0.189 193 666 290 640 556 340 235 206 656 × 2 = 0 + 0.378 387 332 581 281 112 680 470 413 312;
  • 23) 0.378 387 332 581 281 112 680 470 413 312 × 2 = 0 + 0.756 774 665 162 562 225 360 940 826 624;
  • 24) 0.756 774 665 162 562 225 360 940 826 624 × 2 = 1 + 0.513 549 330 325 124 450 721 881 653 248;

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).


4. 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:


0.858 648 867 222 626 824 684 408 350 103(10) =

0.1101 1011 1101 0000 0110 1001(2)

5. Positive number before normalization:

2.858 648 867 222 626 824 684 408 350 103(10) =

10.1101 1011 1101 0000 0110 1001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.858 648 867 222 626 824 684 408 350 103(10) =

10.1101 1011 1101 0000 0110 1001(2) =

10.1101 1011 1101 0000 0110 1001(2) × 20 =

1.0110 1101 1110 1000 0011 0100 1(2) × 21


7. 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 0 (a positive number)

Exponent (unadjusted): 1

Mantissa (not normalized):
1.0110 1101 1110 1000 0011 0100 1


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

1 + 2(8-1) - 1 =

(1 + 127)(10) =

128(10)


9. 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;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

128(10) =

1000 0000(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 011 0110 1111 0100 0001 1010 01 =

011 0110 1111 0100 0001 1010


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
1000 0000

Mantissa (23 bits) =
011 0110 1111 0100 0001 1010


Decimal number 2.858 648 867 222 626 824 684 408 350 103 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0000 - 011 0110 1111 0100 0001 1010

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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